Notes for 4H Galois Theory 2001–2 Andrew...

Notes for 4H Galois Theory 2001–2

Andrew Baker[29/01/2002]

Department of Mathematics, University of Glasgow, Glasgow G12 8QW,

Scotland.

E-mail address: [email protected]: http://www.maths.gla.ac.uk/∼ajb

Bibliography

[1] E. Artin, Galois Theory, Dover Publications (1998); ISBN 0 486 62342 4.

[2] J. B. Fraleigh, A First Course in Abstract Algebra, Addison Wesley (1999); ISBN 0 201 33596 4. [Highly

recommended ]

[3] S. Lang, Algebra, Addison Wesley (1993); ISBN 0 201 55540 9.

[4] R. Lidl & H. Niederreiter, Finite fields, Cambridge University Press (1997); ISBN 0 521 39231 4.

[5] J. Rotman, Galois Theory, Springer-Verlag (1998); ISBN 0 387 98541 7.

[6] I. Stewart, Galois Theory, Chapman and Hall (1989); ISBN 0 412 34550-1.

iii

Contents

Bibliography iii

Chapter 1. Integral domains, fields and polynomial rings 11.1. Recollections on integral domains and fields 11.2. Polynomial rings 51.3. Identifying irreducible polynomials 101.4. Automorphism of rings 13Exercises on Chapter 1 18

Chapter 2. Fields and their extensions 212.1. Fields and subfields 212.2. Simple and finitely generated extensions 23Exercises on Chapter 2 26

Chapter 3. Algebraic extensions of fields 273.1. Algebraic extensions 273.2. Kronecker’s Theorem 303.3. Monomorphisms between extensions 333.4. Algebraic closures 353.5. Multiplicity of roots and separability 383.6. The Primitive Element Theorem 423.7. Normal extensions 43Exercises on Chapter 3 45

Chapter 4. Galois extensions and the Galois Correspondence 474.1. Galois extensions 474.2. Working with Galois groups 484.3. Subgroups of Galois groups and their fixed fields 504.4. Subfields of Galois extensions and relative Galois groups 514.5. The Galois Correspondence 524.6. Kaplansky’s Theorem 534.7. Galois extensions inside the complex numbers 56Exercises on Chapter 4 58

Chapter 5. Galois extensions in positive characteristic 595.1. Finite fields 595.2. Galois groups of finite fields 625.3. The trace and norm mappings 64Exercises on Chapter 5 66

Chapter 6. A Galois Miscellany 676.1. A proof of the Fundamental Theorem of Algebra 676.2. Cyclotomic extensions 676.3. Artin’s Theorem on linear independence of characters 71

v

vi CONTENTS

6.4. Radical extensions 736.5. Solvability and radical extensions 756.6. Symmetric functions 776.7. Galois groups of even and odd permutations 78Exercises on Chapter 6 81

Solutions 83Solutions for exercises on Chapter 1 83Solutions for exercises on Chapter 2 86Solutions for exercises on Chapter 3 88Solutions for exercises on Chapter 4 89Solutions for exercises on Chapter 5 91Solutions for exercises on Chapter 6 92

CHAPTER 1

Integral domains, fields and polynomial rings

A ring will always mean a ring with unity and will usually be commutative. A ring homo-morphism R −→ S is

• a monomorphism if it is injective,• an epimorphism if it is surjective,• an isomorphism if it is both injective and surjective and hence it is invertible with

inverse an isomorphism.

1.1. Recollections on integral domains and fields

The material in this section is standard and much of it should be familiar; see [2, 3, 5] fordetails.

Definition 1.1. A commutative ring R in which there are no zero-divisors is called anintegral domain or an entire ring. This means that for u, v ∈ R,

uv = 0 =⇒ u = 0 or v = 0.

Example 1.2. The following rings are integral domains.

i) The ring of integers, Z.ii) If p is a prime, the ring of integers modulo p, Fp = Z/p = Z/(p).iii) The rings of rational numbers, Q, real numbers, R, and complex numbers, C.

Let I / R be a proper ideal in a commutative ring R, i.e., I 6= R or equivalently I R.Then I is a prime ideal in R if for u, v ∈ R,

uv ∈ I =⇒ u ∈ I or v ∈ I.

Similarly, I is a maximal ideal in R if whenever J / R is a proper ideal and I ⊆ J then J = I.An ideal I / R is principal if

I = (p) = {rp : r ∈ R}for some p ∈ R. Notice that if p, q ∈ R, then (q) = (p) if and only if q = up for some unit u ∈ R.We also write p | x if x ∈ (p).

The following important example should be familiar.

Example 1.3. Every ideal I /Z is principal, so I = (n) for some n ∈ Z which we can alwaystake to be non-negative, n > 0.

An element p ∈ R is prime if (p) / R is a prime ideal; this is equivalent to the requirementthat whenever p | xy with x, y ∈ R then p | x or p | y.

Proposition 1.4. Let R be a commutative ring and I / R and ideal.

i) The quotient ring R/I is an integral domain if and only if I is a prime ideal.ii) The quotient ring R/I is a field if and only if I is a maximal ideal.

Example 1.5. If n > 0, the quotient ring Z/n = Z/(n) is an integral domain if and only ifn is a prime.

1

2 1. INTEGRAL DOMAINS, FIELDS AND POLYNOMIAL RINGS

For any (not necessarily commutative) ring with unit there is an important ring homomor-phism η : Z −→ R called the unit homomorphism and defined by

η(n) = n1 =

1 + · · ·+ 1︸︷︷︸n

if n > 0,

−(1 + · · ·+ 1︸︷︷︸−n

) if n < 0,

0 if n = 0.

Since the unit 1 ∈ R is non-zero, ker η / Z is a non-zero ideal and using the IsomorphismTheorems there is a quotient monomorphism η : Z/ ker η −→ R which allows us to identify thequotient ring Z/ ker η with the image ηZ ⊆ R as a subring of R. Using Example 1.3, we seethat there is a unique non-negative integer p > 0 such that ker η = (p); p is often called thecharacteristic of R and denoted charR.

Lemma 1.6. If R is an integral domain, the characteristic is a prime.

Proof. Consider p = charR. If p = 0 we are done. So suppose that p > 0. The quotientmonomorphism η : Z/ ker η −→ R identifies Z/ ker η with a subring of an integral domain, andany such subring is itself an integral domain, we see that Z/ ker η is an integral domain. Nowby Proposition 1.4(i), ker η = (p) is prime, so by Example 1.5, p is a prime. ¤

Remark 1.7. When discussing a ring with unit R, we can consider it as containing as asubring Z/(charR). In particular, every integral domain contains as a subring either Z = Z/(0)(if charR = 0) or Fp = Z/(p) if p = charR > 0 is a non-zero prime. This subring is sometimescalled the characteristic subring of R. The rings Z and Z/n = Z/(n) for n > 0 are often calledcore rings. When considering integral domains, the rings Z and Fp = Z/p = Z/(p) for p > 0 aprime are called prime rings.

Here is a useful fact about rings containing a finite prime ring.

Proposition 1.8 (Idiot’s Binomial Theorem). Let R be a commutative ring with unit con-taining Fp for some prime p > 0. If u, v ∈ R, then

(u + v)p = up + vp.

Proof. We have p1 = 0 in R. Now the Binomial Theorem gives

(u + v)p = up +(

p

1

)up−1v +

(p

2

)up−2v2 + · · ·+

(p

p− 1

)uvp−1 + vp.

Notice that for 1 6 j 6 p− 1 we have(

p

j

)=

p (p− 1)!j! (p− j)!

.

There are no factors of p appearing in (p− 1)!, j! or (p− j)!, so since this number is an integer

it must be divisible by p. Hence in R we have(

p

j

)1 = 0 which implies the required equation

(u + v)p = up + vp. ¤

A commutative ring with unit k is a field if every non-zero element u ∈ k is a unit in k i.e.,there is an element v ∈ k such that uv = vu = 1. We usually write u−1 for this element v whichis necessarily unique and called the multiplicative inverse of u in k.

Example 1.9. If n > 0, the quotient ring Z/n is a field if and only if n is a positive prime.

Proposition 1.10. If k is a field then it is an integral domain.

1.1. RECOLLECTIONS ON INTEGRAL DOMAINS AND FIELDS 3

Proof. Suppose that u, v ∈ k and uv = 0. Then if u 6= 0 we can multiply this equation byu−1 to obtain

v = u−1uv = 0,

hence v = 0. So at least one of u, v must be 0. ¤

A non-zero element p ∈ R is irreducible if whenever p = uv with u, v ∈ R, either u or v is aunit.

Lemma 1.11. Let R be an integral domain. If p ∈ R is a non-zero prime then it is anirreducible.

Proof. Suppose that p = uv for some u, v ∈ R. Then p | u or p | v, and we might as wellassume that u = tp for some t ∈ R. Then (1− tv)p = 0 and so tv = 1, showing that v is a unitwith inverse t. ¤

Now let D be an integral domain. A natural question to ask is whether D can be found asa subring of a field. This is certainly true for the integers Z which are contained in the field ofrational numbers Q, and for a prime p > 0, the prime ring Fp is itself a field.

Definition 1.12. The fields Q and Fp where p > 0 is prime are the prime fields.

Of course, we can view Z as a subring of any subfield of the complex numbers so an answerto this question may not be unique! However, there is a ‘smallest’ such field which is unique upto an isomorphism.

Theorem 1.13. Let D be an integral domain.i) There is a field of fractions of D, Fr(D), which contains D as a subring.ii) If ϕ : D −→ F is a ring monomorphism into a field F , there is a unique homomorphism

ϕ : Fr(D) −→ F such that ϕ(t) = ϕ(t) for all t ∈ D ⊆ Fr(D).

D

inc²²

ϕ// F

Fr(D)∃! ϕ

<<

Proof. (i) Consider the set

P(D) = {(a, b) : a, b ∈ D, b 6= 0}.Now introduce an equivalence relation ∼ on P(D), namely

(a′, b′) ∼ (a, b) ⇐⇒ ab′ = a′b.

Of course it is necessary to check that this relation is an equivalence relation. We denote theequivalence class of (a, b) by [a, b] and the set of equivalence classes by Fr(D).

Now define addition and multiplication on Fr(D) by

[a, b] + [c, d] = [ad + bc, bd], [a, b][c, d] = [ac, bd].

Again we need to verify that these operations are well defined. For example, if [a′, b′] = [a, b]and [c′, d′] = [c, d], then

(a′d′ + b′c′)bd = a′d′bd + b′c′bd = ab′d′d + b′bcd′ = (ad + bc)b′d′,

and so (a′d′ + b′c′, b′d′) ∼ (ad + bc, bd), hence addition is well defined. A similar calculationshows that (a′c′, b′d′) ∼ (ac, bd), so multiplication is also well defined. It is now straightforwardto show that Fr(D) is a commutative ring with zero 0 = [0, 1] and unit 1 = [1, 1]. In fact, Fr(D)is a field as we shall soon see.


Let [a, b] ∈ Fr(D). Then [a, b] = [0, 1] if and only if (0, 1) ∼ (a, b) which is equivalent torequiring that a = 0; notice that for any b 6= 0, [0, b] = [0, 1]. We also have [a, b] = [1, 1] if andonly if a = b.

Now let [a, b] ∈ Fr(D) be non-zero, i.e., a 6= 0. Then

[a, b][b, a] = [ab, ba] = [1, 1] = 1,

so [a, b] has [b, a] as an inverse, and therefore Fr(D) is a field.We can view D as a subring of Fr(D) using the map

j : D −→ Fr(D); j(t) = [t, 1]

which is a ring homomorphism and actually a monomorphism. Therefore we may identify t ∈ D

with [t, 1] ∈ Fr(D).(ii) First we define a function

Φ: P(D) −→ F ; Φ(a, b) = ϕ(a)ϕ(b)−1.

If (a′, b′) ∼ (a, b) then

Φ(a′, b′) = ϕ(a′)ϕ(b′)−1 = ϕ(a′)ϕ(b)ϕ(b)−1ϕ(b′)−1

= ϕ(a′b)ϕ(b)−1ϕ(b′)−1

= ϕ(ab′)ϕ(b′)−1ϕ(b)−1

= ϕ(a)ϕ(b′)ϕ(b′)−1ϕ(b)−1

= ϕ(a)ϕ(b)−1 = Φ(a, b),

so Φ is constant on each equivalence class of ∼. Hence we can define the function

ϕ : Fr(D) −→ F ; ϕ([a, b]) = Φ(a, b).

It is straightforward to verify that ϕ is a ring homomorphism which agrees with ϕ on the subringD ⊆ Fr(D). ¤

The next three corollaries are left as an exercise. Corollary 1.16 is sometimes said to implythat the construction of Fr(D) is functorial in the integral domain D.

Corollary 1.14. If F is a field then F = Fr(F ).

Corollary 1.15. If D is a subring of a field F , then Fr(D) ⊆ Fr(F ) = F is the smallestsubfield of F containing D.

Corollary 1.16. If D1 and D2 are integral domains and ϕ : D1 −→ D2 a ring monomor-phism, there is a unique induced ring homomorphism ϕ∗ : Fr(D1) −→ Fr(D2) for which ϕ∗(t) =ϕ(t) if t ∈ D1 ⊆ Fr(D1).

D1

ϕ//

inc²²

D2

inc²²

Fr(D1) ϕ∗// Fr(D2)

Moreover, this construction has the properties

• if ϕ : D1 −→ D2 and θ : D2 −→ D3 are monomorphisms between integral domains thenθ∗ ◦ ϕ∗ = (θ ◦ ϕ)∗ as homomorphisms Fr(D1) −→ Fr(D3);

• for any integral domain D, the identity homomorphism id : D −→ D induces the iden-tity homomorphism (id)∗ = id: Fr(D) −→ Fr(D).

1.2. POLYNOMIAL RINGS 5

D1

ϕ//

inc²²

D2θ //

inc²²

D3

inc²²

Fr(D1) ϕ∗// Fr(D2)

θ∗// Fr(D3)

Did //

inc²²

D

inc²²

Fr(D)id∗ = id

// Fr(D2)

Remark 1.17. When working with a field of fractions it is usual to adopt the notationa

b= [a, b]

for the equivalence class of (a, b). The rules for algebraic manipulation of such symbols are theusual ones for working with fractions, i.e.,

a1

b1+

a2

b2=

a1b2 + a2b1

b1b2,

a1

b1× a2

b2=

a1

b1

a2

b2=

a1a2

b1b2.

We also mention that the field of fractions of an integral domain is sometimes called its field ofquotients, however as the word quotient is also associated to quotient rings we prefer to avoidusing that terminology.

1.2. Polynomial rings

Let R be a commutative ring. We will make frequent use of the ring R[X] of polynomialsover R in an indeterminate X. This consists of elements of form

p(X) = p0 + p1X + · · ·+ pmXm

where m > 0 and p0, p1, . . . , pm ∈ R; such p(X) are called polynomials. Addition and multipli-cation in R[X] are defined by

(p0 + p1X + · · ·+ pmXm) + (q0 + q1X + · · ·+ qmXm) =

(p0 + q0) + (p1 + q1)X + · · ·+ (pm + qm)Xm),

and

(p0 + p1X + · · ·+ pmXm)(q0 + q1X + · · ·+ qmXm) =

(p0q0) + (p0q1 + p1q0)X + · · ·+ (p0qm + p1qm−1 + · · ·+ pm−1q1 + pmq0)X2m.

Then R[X] is a commutative ring with the constant polynomials 0 and 1 as its zero and unit.We identify r ∈ R with the obvious constant polynomial; this allows us to view R as a subringof R[X] and the inclusion function inc : R −→ R[X] is a monomorphism.

More generally, we inductively can define the ring of polynomials in n indeterminatesX1, . . . , Xn over R,

R[X1, . . . , Xn] = R[X1, . . . , Xn−1][Xn]

for n > 1. Again there is an inclusion monomorphism inc: R −→ R[X1, . . . , Xn] which sendseach element of R to the corresponding constant polynomial.

These polynomial rings have an important universal property.

Theorem 1.18 (Homomorphism Extension Property). Let ϕ : R −→ S be a ring homomor-phism.

i) For each s ∈ S there is a unique ring homomorphism ϕs : R[X] −→ S for which• ϕs(r) = ϕ(r) for all r ∈ R,• ϕs(X) = s.


R

inc²²

ϕ// S

R[X]∃! ϕs

==

ii) For n > 1 and s1, . . . , sn ∈ S, there is a unique ring homomorphism

ϕs1,...,sn : R[X1, . . . , Xn] −→ S

for which• ϕs1,...,sn(r) = ϕ(r) for all r ∈ R,• ϕs1,...,sn(Xi) = si for i = 1, . . . , n.

R

inc²²

ϕ// S

R[X1, . . . , Xn]∃! ϕs1,...,sn

88

Sketch of the proof. (i) For a polynomial p(X) = p0 + p1X + · · ·+ pmXm ∈ R[X], wedefine

(1.1) ϕs(p(X)) = p0 + p1s + · · ·+ pmsm ∈ S.

It is then straightforward to check that ϕs is a ring homomorphism with the stated propertiesand moreover is the unique such homomorphism.(ii) is proved by induction on n using (i). ¤

We will refer to ϕs1,...,sn as the extension of ϕ by evaluation at s1, . . . , sn. It is standard towrite

p(s1, . . . , sn) = ϕs1,...,sn(p(X1, . . . , Xn)).

An extremely important special case occurs when we start with the identity homomorphismid: R −→ R and r1, . . . , rn ∈ R; then we have the homomorphism

εr1,...,rn = idr1,...,rn : R[X1, . . . , Xn] −→ R.

Slightly more generally we may take the inclusion of a subring inc : R −→ S and s1, . . . , sn ∈ S;then

εs1,...,sn = incs1,...,sn : R[X1, . . . , Xn] −→ S

is called evaluation at s1, . . . , sn and we denote its image by

R[s1, . . . , sn] = εs1,...,sn R[X1, . . . , Xn] ⊆ S.

Then R[s1, . . . , sn] is a subring of S, called the subring generated by s1, . . . , sn over R.Here is an example illustrating how we will use such evaluation homomorphisms.

Example 1.19. Consider the inclusion homomorphism inc: Q −→ C. The we have theevaluation at i homomorphism εi, for which εi(X) = i. We easily find that εiQ[X] ⊆ C is asubring Q[i] ⊆ C consisting of the complex numbers of form a + bi with a, b ∈ Q.

Notice that if we instead used −i instead of i, evaluation at −i, ε−i, would also giveε−iQ[X] = Q[i]. These evaluation homomorphisms are related by complex conjugation since

ε−i(p(X)) = εi(p(X)),

which is equivalent to the functional equation

ε−i = ( ) ◦ εi .


Notice also that in these examples we have

ker ε−i = ker εi = (X2 + 1) /Q[X],

hence we also haveQ[i] ∼= Q[X]/(X2 + 1).

In fact (X2 + 1) is actually a maximal ideal and so Q[i] ⊆ C is a subfield; we later we will writeQ(i) for this.

Proposition 1.20. Let R be an integral domain.i) The ring R[X] of polynomials in an indeterminate X over R is an integral domain.ii) The ring R[X1, . . . , Xn] of polynomials in n indeterminates X1, . . . , Xn over R is an

integral domain.

Corollary 1.21. If k is a field and n > 1, the polynomial ring k[X1, . . . , Xn] in n indeter-minates X1, . . . , Xn is an integral domain.

As we will make considerable use of such rings we next give some of their important prop-erties. First we recall the Long Division property of a polynomial ring in one indeterminate X

over a field. The following is a basic property that we will not prove but is to be found in abasic course on commutative rings or any introductory book on this subject.

Theorem 1.22. Let k be a field and X an indeterminate. Let f(X), d(X) ∈ k[X] and assumethat d(X) 6= 0 so that deg d(X) > 0. Then there are unique polynomials q(X), r(X) ∈ k[X] forwhich f(X) = q(X)d(X) + r(X) with deg r(X) < d or r(X) = 0.

In the situation discussed in this result, the following names are often used. We refer to theprocess of finding q(X) and r(X) as long division of f(X) by d(X). Also,

f(X) = the dividend , d(X) = the divisor , q(X) = the quotient , r(X) = the remainder .

Example 1.23. If k = Q, find the quotient and remainder when the long division of f(X) =6X4 − 6X3 + 3X2 − 3X + 1 by d(X) = 2X2 + 1 is performed.

Solution. In the usual notation we have the following calculation.

3X2 − 3X

2X2 + 1 | 6X4 − 6X3 + 3X2 − 3X + 1

6X4 + 0X3 + 3X2 + 0X + 0

− 6X3 + 0X2 − 3X + 1

− 6X3 + 0X2 − 3X + 0

1

Hence6X4 − 6X3 + 3X2 − 3X + 1 = (3X2 − 3X)(2X2 + 1) + 1,

giving q(X) = 3X2 − 3X and r(X) = 1. ¤

Example 1.24. If k = F5, find the quotient and remainder when long division of f(X) =10X5 + 6X4 − 6X3 + 3X2 − 3X + 1 by d(X) = 2X2 + 1 is performed.

Solution. First notice that working modulo 5 we have

f(X) = 10X5 + 6X4 − 6X3 + 3X2 − 3X + 1 ≡ X4 + 4X3 + 3X2 + 2X + 1 mod 5.

Notice also that multiplicative inverses in F5 are given by

2−1 ≡ 3 mod 5, 3−1 ≡ 2 mod 5, 4−1 ≡ 4 (mod 5).


We have the following calculation.

3X2 + 2X

2X2 + 1 | 6X4 + 4X3 + 3X2 + 2X + 1

6X4 + 0X3 + 3X2 + 0X + 0

4X3 + 0X2 + 2X + 1

4X3 + 0X2 + 2X + 0

1

Hence6X4 − 6X3 + 3X2 − 3X + 1 ≡ (3X2 + 2X)(2X2 + 1) + 1 mod 5,

giving q(X) = 3X2 + 2X and r(X) = 1. ¤

An important consequence of Theorem 1.22 is the following which uses the Euclidean Algo-rithm.

Corollary 1.25. Let k be a field and X an indeterminate. Let f(X), g(X) ∈ k[X] benon-zero. Then there are a(X), b(X) ∈ k[X] such that

a(X)f(X) + b(X)g(X) = gcd(f(X), g(X)).

Here the greatest common divisor gcd(f(X), g(X)) of f(X), g(X) is the monic polynomialof greatest degree which divides both of f(X), g(X).

Proposition 1.26. Let k be a field and X an indeterminate. Then a non-constant polyno-mial p(X) ∈ k[X] is an irreducible if and only if it is a prime.

Proof. By Lemma 1.11 we already know that p(X) is irreducible if it is prime. So supposethat p(X) is irreducible and that p(X) | u(X)v(X) for u(X), v(X) ∈ k[X].

Then by Corollary 1.25, there are a(X), b(X) ∈ k[X] such that

a(X)p(X) + b(X)u(X) = gcd(p(X), u(X)).

But since p(X) is irreducible, gcd(p(X), u(X)) = 1, hence

a(X)p(X) + b(X)u(X) = 1.

Multiplying through by v(X) gives

a(X)p(X)v(X) + b(X)u(X)v(X) = v(X)

and so p(X) | v(X). This shows that p(X) | u(X) or p(X) | v(X) and so p(X) is prime. ¤

Theorem 1.27. Let k be a field and X an indeterminate.

i) Every ideal I / k[X] is principal, i.e., I = (h(X)) for some h(X) ∈ k[X].ii) The ideal (p(X))/k[X] is prime if and only if p(X) = 0 or p(X) is irreducible in k[X].iii) The quotient ring k[X]/(p(X)) is an integral domain if and only if p(X) = 0 or p(X)

is irreducible in k[X].iv) The quotient ring k[X]/(p(X)) is a field if and only if p(X) is an irreducible in k[X].

Proof. (i) Let I / k[X] and assume that I 6= (0). Then there must be at least one elementof I with positive degree and so we can choose h(X) ∈ I of minimal degree, say d = deg h(X).

Now let p(X) ∈ I. By Long Division, there are q(X), r(X) ∈ k[X] such that

p(X) = q(X)h(X) + r(X) and deg r(X) < d or r(X) = 0.


Since p(X) and h(X) are in the ideal I, we also have

r(X) = p(X)− q(X)h(X) ∈ I.

If r(X) 6= 0 this would contradict the minimality of d, so we must have r(X) = 0, showing thatp(X) = q(X)h(X). Thus I ⊆ (p(X)) ⊆ I and therefore I = (p(X)).(ii) This follows from Proposition 1.26.(iii) This follows from Proposition 1.4(i).(iv) Since k[X] is an integral domain and not a field, it is follows that if k[X]/(p(X)) is a fieldthen because it is an integral domain, p(X) is an irreducible by (iii).

Suppose that p(X) is irreducible (and hence is non-zero). Then for any q(X) ∈ k[X] withq(X) /∈ (p(X)) we can use the Euclidean Algorithm of Corollary 1.25 to write

a(X)p(X) + b(X)q(X) = gcd(p(X), q(X)).

But gcd(p(X), q(X)) = 1 since p(X) is irreducible, so

a(X)p(X) + b(X)q(X) = 1.

This shows that in the quotient ring k[X]/(p(X)) the residue class of q(X) has the residue classof b(X) as its inverse. ¤

Remark 1.28. In connection with Theorem 1.27(i), notice that if p(X) ∈ k[X], then pro-vided d = deg p(X) > 0, we have for some pd 6= 0,

p(X) = p0 + p1X + · · ·+ pdXd = pdq(X),

where

q(X) = p−1d p0 + p−1

d p1X + · · ·+ p−1d pd−1X

d−1 + Xd.

This easily implies that as ideals of k[X], (p(X)) = (q(X)). So we can always find a monicpolynomial as the generator of a given ideal, and this is actually unique.

Proposition 1.29 (Unique Factorization Property of k[X]). Every non-constant polynomialf(x) ∈ k[X] has a factorization

f(x) = cp1(X) · · · pk(X),

where c ∈ k, and p1(X), . . . , pk(X) ∈ k[X] are irreducible monic polynomials. Moreover, c isunique and the sequence of polynomials p1(X), . . . , pk(X) is unique apart from the order of theterms.

Sketch proof. The existence is proved by induction on the degree of f(X) and beginswith the obvious case deg f(X) = 1. If deg f(X) > 1, then either f(X) is already irreducible, orf(X) = f1(X)f2(X) with both factors of positive degree, and therefore deg fj(X) < deg f(X).This gives the inductive step.

To prove uniqueness, suppose that

p1(X) · · · pk(X) = q1(X) · · · q`(X)

where pi(X), qj(X) ∈ k[X] are irreducible monic polynomials. Then by Proposition 1.26, eachpi(X) is prime, hence divides one of the qj(X), hence must equal it. By reordering we canassume that pi(X) = qi(X) and k 6 `. Then after cancelling common factors we obtain

qk+1(X) · · · q`(X) = 1,

hence k = `. ¤


Corollary 1.30. Suppose that f(X) ∈ k[X] factors into linear factors

f(X) = c(X − u1) · · · (X − ud),

where u1, . . . , ud ∈ k. Then the sequence of roots u1, . . . , ud is unique apart from the order. Inparticular, if v1, . . . , vr are the distinct roots, then

f(X) = c(X − v1)m1 · · · (X − vr)mr ,

where mi > 0 and this factorization is unique apart from the order of the pairs (vi,mi).

Corollary 1.31. The number of distinct roots of the non-constant polynomial f(X) ∈ k[X]is at most deg f(X).

Definition 1.32. If k is a field and X an indeterminate, then the ring of rational functionsk(X) is the fraction field of k[X]. The elements of k(X) are fractions

a0 + a1X + · · ·+ amXm

b0 + b1X + · · ·+ bnXn

with ai, bj ∈ k.

1.3. Identifying irreducible polynomials

We will need some effective methods for identifying when a polynomial is irreducible in apolynomial ring k[X] where k is a field.

Let us consider factorisation of polynomials overQ. If f(X) ∈ Z[X] then we can also considerf(X) as an element of Q[X]. If R = Z or Q, we say that f(X) has a proper factorisation overR if f(X) = g(X)h(X) for some g(X), h(X) ∈ R[X] with deg g(X) > 0 and deg h(X) > 0.

Proposition 1.33 (Gauss’s Lemma). Let f(X) ∈ Z[X]. Then f(X) has a proper factori-sation over Z if and only it has a proper factorisation over Q.

So to find factors of f(X) it is sufficient to look only in Z[X].The next result is a special case of the Eisenstein Irreducibility Criterion.

Proposition 1.34 (Eisenstein Criterion). Let

f(X) = a0 + a1X + · · ·+ ad−1Xd−1 + adX

d ∈ Z[X]

and d = deg f(X). Suppose that p > 0 is a prime in Z for which the following conditions hold:

• ak ≡ 0 mod p for k = 0, . . . , d− 1;• a0 6≡ 0 mod p2;• ad 6≡ 0 mod p.

Then f(X) is irreducible in Q[X] and hence also in Z[X].

Example 1.35. Let p > 2 be a prime. Then the polynomial

Φp(X) = 1 + X + · · ·+ Xp−1 ∈ Z[X]

is irreducible in Q[X] and hence also in Z[X].

Proof. For an indeterminate Y and working in Z[Y ],

Φp(Y + 1)Y = ((Y + 1)− 1)(1 + (Y + 1) + · · ·+ (Y + 1)p−1)

= (Y + 1)p − 1

=p∑

k=1

(p

k

)(Y + 1)k

≡ (Y + 1)p mod p,

1.3. IDENTIFYING IRREDUCIBLE POLYNOMIALS 11

since p divides the numerator of (p

k

)=

p!k! (p− k)!

when k = 1, . . . , p− 1. Also, (p

1

)= p 6≡ 0 mod p2.

This shows that Φp(Y + 1) is irreducible in Z[Y ] and it is easy to see that this implies thatΦp(X) is irreducible in Z[X]. ¤

Example 1.36. As examples we have the irreducible polynomials

Φ2(X) = 1 + X,

Φ3(X) = 1 + X + X2,

Φ5(X) = 1 + X + X2 + X3 + X4,

Φ7(X) = 1 + X + X2 + X3 + X4 + X5 + X6,

Φ11(X) = 1 + X + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10.

These are examples of the cyclotomic polynomials Φn(X) ∈ Z[X]; they are defined for alln > 1 and satisfy

(1.2) Xn − 1 =∏

d|nΦd(X),

where the product is taken over all the positive divisors of n. For example,

X2 − 1 = (X − 1)(X + 1) = Φ1(X)Φ2(X),

X3 − 1 = (X − 1)(X2 + X + 1) = Φ1(X)Φ3(X),

X4 − 1 = (X − 1)(X + 1)(X2 + 1) = Φ1(X)Φ2(X)Φ4(X),

X5 − 1 = (X − 1)(X4 + X3 + X + 1) = Φ1(X)Φ5(X),

X6 − 1 = (X − 1)(X + 1)(X2 + X + 1)(X2 −X + 1) = Φ1(X)Φ2(X)Φ3(X)Φ6(X),

X12 − 1 = (X − 1)(X + 1)(X2 + X + 1)(X2 + 1)(X2 −X + 1)(X4 −X2 + 1)

= Φ1(X)Φ2(X)Φ3(X)Φ4(X)Φ6(X)Φ12(X).

Cyclotomic polynomials can be computed recursively with the aid of Equation (1.2). If we knowΦk(X) for k < n, then

(1.3) Φn(X) =Xn − 1∏

d|nd<n

Φd(X).

The degree of Φn(X) involves a function of n probably familiar from elementary NumberTheory.

Definition 1.37. The Euler function ϕ : N −→ N is defined by

ϕ(n) = number of k = 1, . . . , n for which gcd(n, k) = 1

= |(Z/n)×| = number of units in Z/n

= number of generators of the cyclic group Z/n.

In particular, if p > 2 is a prime then ϕ(p) = p− 1. Of course, ϕ(1) = 1.


It can be shown that for each natural number n,

(1.4)∑

d|nϕ(d) = n.

Notice that we can inductively determine ϕ(n) using this equation. For example, if p and q aredistinct primes, then

ϕ(pq) = pq − (ϕ(p) + ϕ(q) + ϕ(1)) = pq − (p− 1)− (q − 1)− 1 = (p− 1)(q − 1).

Recall that an element ζ of a field K is a primitive n-th root of unity if

min{k : 1 6 k and ζk = 1} = n.

We think of ζn = e2πi/n as the standard complex primitive n-th root of unity. Then everycomplex n-th root of unity has the form ζk

n = e2πik/n for k = 0, 1, . . . , n− 1.

Theorem 1.38. For each n > 1, the cyclotomic polynomial Φn(X) is irreducible in Q[X]and hence in Z[X]. The complex roots of Φn(X) are the primitive n-th roots of unity,

ζkn = e2πik/n (0 6 k 6 n− 1, gcd(k, n) = 1).

and the number of these is deg Φn(X) = ϕ(n). Hence,

Φn(X) =∏

t=1,...,n−1gcd(t,n)=1

(X − ζtn).

Proof. We will give a reformulation and proof of this in Theorem 6.2. ¤

Example 1.39. For n = 6 we have

ζ6 = e2πi/6 = eπi/3 =12

+√

32

i.

Then ϕ(6) = 2 andΦ6(X) = X2 −X + 1 = (X − ζ6)(X − ζ5

6 ).

It is also worth recording a related general result on cyclic groups.

Proposition 1.40. Let n > 1 and C = 〈g〉 be a cyclic group of order n and a generator g.Then an element gr ∈ C is a generator if and only if gcd(r, n) = 1; the number of such elementsof C is ϕ(n).

This leads to a useful group theoretic result.

Lemma 1.41. Let G be a finite group satisfying the following condition:

• For each n > 1, there are at most n solutions of xn = ι in G.

Then G is cyclic and in particular is abelian.

Proof. Let θG(d) denote the number of elements in G of order d. By Lagrange’s Theorem,θG(d) = 0 unless d divides |G|. Since

G =⋃

d||G|{g ∈ G : |g| = d},

we have|G| =

∑

d||G|θG(d).

Recall the Euler ϕ-function satisfies Equation (1.4), hence

|G| =∑

d||G|ϕ(d).

1.4. AUTOMORPHISM OF RINGS 13

Combining these we obtain

(1.5)∑

d||G|θG(d) =

∑

d||G|ϕ(d).

Let d be a divisor of |G|. By Proposition 1.40, for each element g ∈ G of order d, the cyclicsubgroup 〈g〉 6 G has ϕ(d) generators, each of order d. As there are at most d such elements g

in G, this gives θG(d) 6 ϕ(d). So we have∑

d||G|θG(d) 6

∑

d||G|ϕ(d).

Now if θG(d) < ϕ(d) for some d, we would have a strict inequality in place of Equation (1.5).Hence θG(d) = ϕ(d) for all d. In particular, there are ϕ(|G|) elements of order |G|, hence theremust be an element of order |G|, so G is cyclic. ¤

The above results for polynomials over Q and Z have analogues over the field of fractionsk(T ) and polynomial ring k[T ], where k is a field.

A polynomial f(X) ∈ k[T ][X] is an element of k(T )[X]. If R = k[T ] or k(T ), we say thatf(X) has a proper factorisation over R if f(X) = g(X)h(X) for some g(X), h(X) ∈ R[X] withdeg g(X) > 0 and deg h(X) > 0.

Proposition 1.42 (Gauss’s Lemma). Let f(X) ∈ k[T ][X]. Then f(X) has a proper fac-torisation over k[T ] if and only it has a proper factorisation over k(T ).

Proposition 1.43 (Eisenstein Criterion). Let

f(X) = a0 + a1X + · · ·+ ad−1Xd−1 + adX

d ∈ k[T ][X]

and d = deg f(X). Suppose that p(T ) ∈ k[T ] is an irreducible for which the following conditionshold:

• ak ≡ 0 mod p(T ) for k = 0, . . . , d− 1;• a0 6≡ 0 mod p(T )2;• ad 6≡ 0 mod p(T ).

Then f(X) is irreducible in k(T )[X] and hence also in k[T ][X].

Example 1.44. Let k be a field. Then the polynomial Xn − T is irreducible in k(T )[X].

1.4. Automorphism of rings

Definition 1.45. Let R be a ring and R0 ⊆ R a subring.• An automorphism of R is a ring isomorphism α : R −→ R. The set of all such auto-

morphisms is denoted Aut(R).• An automorphism of R over R0 is a ring isomorphism α : R −→ R for which α(r) = r

whenever r ∈ R0. The set of all such automorphisms of R over R0 denoted AutR0(R).

Proposition 1.46. For a ring R with a subring R0 ⊆ R, Aut(R) and AutR0(R) form groupsunder composition of functions.

Proof. The composition α◦β of two automorphisms α, β : R −→ R is also an automorphismof R as is the inverse of α. The identity function id = idR : R −→ R is an automorphism. HenceAut(R) forms a group under composition. The argument for AutR0(R) is similar. ¤

Proposition 1.47. Let R be one of the core rings Z or Z/n with n > 1. Theni) The only automorphism of R is the identity, i.e., Aut(R) = {id}.ii) If S is a ring containing a core ring R and α ∈ Aut(S), then α restricts to the identity

on R, i.e., α(r) = r for all r ∈ R. Hence, Aut(S) = AutR(S).


Proof. (i) For such a core ring R, every element has the form k1 for some k ∈ Z. For anautomorphism α of R,

α(k1) =

α(1) + · · ·+ α(1)︸︷︷︸k

if k > 0,

−(α(1) + · · ·+ α(1)︸︷︷︸−k

) if k < 0,

α(0) if k = 0

=

1 + · · ·+ 1︸︷︷︸k

if k > 0,

−(1 + · · ·+ 1︸︷︷︸−k

) if k < 0,

0 if k = 0

=k1.

Thus α = id.(ii) For α ∈ Aut(S), α(1) = 1 and essentially the same argument as in (i) shows that α(r) = r

for all r ∈ R. ¤

Proposition 1.48. Let D be an integral domain and α : D −→ D be an automorphism.Then the induced homomorphism gives an automorphism α∗ : Fr(D) −→ Fr(D).

Proof. Given α, the induced homomorphism α∗ : Fr(D) −→ Fr(D) exists and we needto show it has an inverse. The inverse automorphism α−1 : D −→ D also gives rise to aninduced homomorphism (α−1)∗ : Fr(D) −→ Fr(D). Since α−1 ◦ α = id = α ◦ α−1, we can applyCorollary 1.16 to show that

(α−1)∗ ◦ (α)∗ = id = (α)∗ ◦ (α−1)∗,

therefore (α)∗ is invertible and has inverse (α−1)∗. ¤

Corollary 1.49. There is a monomorphism of groups

( )∗ : Aut(D) −→ Aut(Fr(D)); α 7−→ α∗.

Note that this monomorphism need not be an epimorphism since it is possible that anautomorphism of Fr(D) could map some elements of D ⊆ Fr(D) out of D.

Example 1.50. The field of fractions of the ring of integers Z is the field of rationals Q.The homomorphism

( )∗ : Aut(Z) −→ Aut(Q); α 7−→ α∗is an isomorphism and hence Aut(Q) = {id}.

Combining this example with Proposition 1.47(ii) we obtain a result which will prove usefullater.

Proposition 1.51. Let k be one of the prime fields Q or Fp with p > 0 prime. If R is aring containing k as a subring, then every automorphism of R restricts to the identity on k, i.e.,Aut(R) = Aut|(R).

Let k be a field. The group of invertible 2 × 2 matrices over k is the 2 × 2 general lineargroup over k,

GL2(k) ={[

a11 a12

a21 a22

]: aij ∈ k, a11a22 − a12a21 6= 0

}


The scalar matrices form a normal subgroup

Scal2(k) = {diag(t, t) : t ∈ k, t 6= 0} / GL2(k).

The quotient group is called the 2× 2 projective general linear group over k,

PGL2(k) = GL2(k)/Scal2(k).

Notice that GL2(k) has another interesting subgroup called the affine subgroup,

Aff1(k) ={[

a b

0 1

]: a, b ∈ k, a 6= 0

}6 GL2(k).

Example 1.52. Let k be a field and X an indeterminate. Then Aut|(k[X]) and henceAut|(k(X)), contains a subgroup isomorphic to Aff1(k). In fact, Aut|(k[X]) ∼= Aff1(k).

Proof. We begin by showing that to each affine matrix

A =[a b

0 1

]∈ Aff1(k)

there is an associated automorphism αA : k[X] −→ k[X].For this we use the element aX + b ∈ k[X] together with the extension result of Theo-

rem 1.18(i) to obtain a homomorphism αA : k[X] −→ k[X] with αA(X) = aX + b. Using theinverse matrix

A−1 =[a−1 −a−1b

0 1

]

we similarly obtain a homomorphism αA−1 : k[X] −→ k[X] for which αA−1(X) = a−1X − a−1b.Using the same line of argument as in the proof of Proposition 1.48 (or doing a direct calculation)we see that αA−1 is the inverse of αA an so αA ∈ Aut|(k[X]). It is straightforward to checkthat for A1, A2 ∈ Aff1(k),

αA2A1 = αA1 ◦ αA2 ,

(note the order!) hence there is a homomorphism of groups

Aff1(k) −→ Aut|(k[X]); A 7−→ αA−1 ,

which is easily seen to be a monomorphism. Composing with ( )∗ we see that there is amonomorphism Aff1(k) −→ Aut|(k(X)). In fact, this is also an epimorphism and we leave theproof of this as an exercise. ¤

Example 1.53. Let k be a field and X an indeterminate. Then Aut|(k(X)) contains asubgroup isomorphic to PGL2(k). In fact, Aut|(k(X)) ∼= PGL2(k).

Proof. We begin by showing that to each invertible matrix

A =[a11 a12

a21 a22

]∈ GL2(k)

there is an associated automorphism αA : k(X) −→ k(X).We begin by choosing the element (a11X + a12)/(a21X + a22) ∈ k(X) and then using Theo-

rem 1.18(i) to obtain a homomorphism k[X] −→ k(X) that sends X to (a11X+a12)/(a21X+a22).By applying ( )∗ to this we obtain a homomorphism (known as a fractional linear transforma-tion) αA : k(X) −→ k(X) for which

αA(X) =a11X + a12

a21X + a22.

Again we find thatαA2A1 = αA1 ◦ αA2 .


There is an associated homomorphism of groups GL2(k) −→ Aut|(k(X)) sending A to αA−1.

However, this is not an injection in general since for each scalar matrix diag(t, t),

αdiag(t,t)(X) =tX

t= X,

showing that αdiag(t,t) is the identity function.In fact it is easy to see that Scal2(k)/GL2(k) is the kernel of this homomorphism. Therefore

passing to the quotient PGL2(k) = GL2(k)/Scal2(k) we obtain a monomorphism PGL2(k) −→Aut|(k(X)). There is one case where Scal2(k) is the trivial group, namely k = F2.

To show that every automorphism of k(X) is a fractional linear transformation is less ele-mentary. We give a sketch proof for the case of k = C; actually this argument can be modifiedto work for any algebraically closed field, but an easy argument then shows the general case.

Let α ∈ AutC(C(X)). There is an associated rational (hence meromorphic) function f givenby z 7−→ f(z), where α(X) = f(X), defined on C with the poles of f deleted. If we write

f(X) =p(X)q(X)

where p(X), q(X) ∈ C[X] have no common factors of positive degree, then we define the orderof f(X) to be

ord f = max{deg p(X),deg q(X)}.Now let c ∈ C. Then the number of solutions counted with algebraic multiplicity of the equationf(z) = c turns out to be ord f . Also, if deg p(X) 6 deg q(X) then the number of poles of f

counted with algebraic multiplicity is also ord f . Finally, if deg p(X) > deg q(X) then we canwrite

f(X) = p1(X) +p0(X)q(X)

,

where p0(X), p1(X) ∈ C[X] and deg p0(X) < deg q(X). Then the number of poles of f countedwith algebraic multiplicity is

deg p1(X) + ordp0

q.

Now it is easy to see that since α is invertible so is the function f . But this can only happenif the function f is injective which means that all of these numbers must be 1, hence ord f = 1.Thus

f(X) =aX + b

cX + d6= constant

and the matrix[a b

c d

]must be invertible. ¤

Clearly not every fractional linear transformation αA : k(X) −→ k(X) maps polynomials topolynomials so ( )∗ : Aut|(k[X]) −→ Aut|(k(X)) is not an epimorphism.

Now we turn to a more familiar field R, the real numbers.

Proposition 1.54. The only automorphism of the field R is the identity function, henceAut(R) = {id}.

Proof. First we note that Q ⊆ R is a subring and if α ∈ Aut(R) then α(q) = q for q ∈ Qby Example 1.50.

We recall from Analysis that the rational numbers are dense in the real numbers in thesense that each r ∈ R can be expressed as a limit r = limn→∞ qn, where qn ∈ Q. Then for acontinuous function f : R −→ R, its value at r depends on its values on Q since

f(r) = f( limn→∞ qn) = lim

n→∞ f(qn).

We will show that an automorphism α ∈ Aut(R) is continuous.


First recall that for x, y ∈ R,

x < y ⇐⇒ 0 < y − x ⇐⇒ y − x = t2 for some non-zero t ∈ R.

Now for α ∈ Aut(R) and s ∈ R, we have α(s2) = α(s)2. Hence,

x < y =⇒ α(y)− α(x) = α(t)2 for some non-zero t ∈ R =⇒ α(x) < α(y).

So α preserves order and fixes rational numbers.Now let x ∈ R and ε > 0. Then we can choose a rational number q such that 0 < q 6 ε.

Taking δ = q we find that for y ∈ R with |y − x| < δ (i.e., −δ < y − x < δ) we have

−δ = α(−δ) < α(y)− α(x) < α(δ) = δ,

hence|α(y)− α(x)| < δ 6 ε.

This shows that α is continuous at x.Thus every automorphism of R is continuous and fixes all rational numbers, hence it must

be the identity function. ¤

Remark 1.55. If we try to determine Aut(C) the answer turns out to be much more com-plicated. It is easy to see that complex conjugation ( ) : C −→ C is an automorphism of Cand fixes every real number, i.e., ( ) ∈ AutR(C); in fact, AutR(C) = {id, ( )}. However, it isnot true that every α ∈ Aut(C) fixes every real number! The automorphism group Aut(C) isactually enormous but it is hard to find an explicit element other than id and ( ). Note thatgiven an automorphism α ∈ Aut(C), the composition α ◦ ( ) ◦ α−1 is also self inverse, so thereare many elements of order 2 in the group Aut(C).


Exercises on Chapter 1

1.1. Let R be an integral domain.

(a) Show that every subring S ⊆ R is also an integral domain. What is the relationshipbetween charS and charR?

(b) If R is a field, give an example to show that a subring of R need not be a field.

1.2. For each of the following rings R, find the characteristic charR and the characteristicsubring of R. Determine which of these rings is an integral domain. In (b) and (c), A is acommutative ring.

(a) Any subring R ⊆ C.(b) The polynomial ring R = A[X].(c) The ring of n× n matrices over A,

R = Matn(A) =

a11 . . . a1n...

. . ....

an1 . . . ann

: aij ∈ A

.

1.3. If R is a commutative ring with unit containing the prime field Fp for some prime p > 0,show that the function ϕ : R −→ R given by ϕ(t) = tp, defines a ring homomorphism. Giveexamples to show that ϕ need not be surjective or injective.

1.4. Let R and S be rings with unity and Q / S a prime ideal.

(a) If ϕ : R −→ S is a ring homomorphism, show that

ϕ−1Q = {r ∈ R : ϕ(r) ∈ Q} ⊆ R

is a prime ideal of R.(b) If R ⊆ S is a subring, show that Q ∩R is a prime ideal of R.(c) If the word ‘prime’ is replaced by ‘maximal’ throughout, are the results in parts (a)

and (b) still true? [Hint: look for a counterexample.](d) If R ⊆ S is a subring and P /R is a maximal ideal, suppose that Q/S is a prime ideal

for which P ⊆ Q. Show that Q ∩R = P .

1.5. Let k be a field, R be a ring with unit and ϕ : k −→ R a ring homomorphism. Show thatϕ is a monomorphism.

1.6. Consider the sets

Z(i) = {u + vi : u, v ∈ Z} ⊆ C, Q(i) = {u + vi : u, v ∈ Q} ⊆ C.

(a) Show that Z(i) and Q(i) are subrings of C. Also show that Z(i) is an integral domain,Q(i) is a field and Z(i) is a subring of Q(i).

(b) Show that the inclusion homomorphism inc: Z(i) −→ Q(i) extends to a monomorphisminc∗ : Fr(Z(i)) −→ Q(i).

(c) Show that inc∗ is an isomorphism, so Fr(Z(i)) = Q(i).

1.7. Let k be a field and k[[X]] be the set consisting of all power series∞∑

k=0

akXk = a0 + a1X + · · ·+ akX

k + · · · ,

with ak ∈ k.

19

(a) Show that this can be made into an integral domain containing k[X] as a subring bydefining addition and multiplication in the obvious way.

(b) Show that∑∞

k=0 akXk ∈ k[[X]] is a unit if and only if a0 6= 0.

(c) Show that Fr(k[[X]]) consists of all finite tailed Laurent series∞∑

k=`

akXk = a`X

` + a`+1X`+1 + · · ·+ akX

k + · · ·

for some ` ∈ Z and a| ∈ k.1.8. Taking k = Q, find the quotient and remainder when performing long division of f(X) =6X4 − 6X3 + 3X2 − 3X − 2 by d(X) = 2X3 + X + 3.

1.9. Taking k = F3, find the quotient and remainder when performing long division of f(X) =2X3 + 2X2 + X + 1 by d(X) = 2X3 + 2X.

1.10. Let p > 0 be a prime. Suppose that f(X) = a0 + a1X + · · · + anXn ∈ Z[X] withp - an and that f(X) ∈ Fp[X] denotes the polynomial obtained by reducing the coefficients off(X) modulo p. If f(X) is irreducible, show that f(X) is irreducible. Which of the followingpolynomials in Z[X] is irreducible?

X3−X +1, X3 +2X +1, X3 +X− 1, X5−X +1, X5 +X− 1, 5X3− 10X +X2− 2.

1.11. Find generators for each of the following ideals:

I1 = {f(X) ∈ Q[X] : f(i) = 0} /Q[X], I2 = {f(X) ∈ Q[X] : f(√

2 i) = 0} /Q[X],

I3 = {f(X) ∈ Q[X] : f(√

2) = 0} /Q[X], I4 = {f(X) ∈ R[X] : f(√

2) = 0} / R[X],

I5 = {f(X) ∈ R[X] : f(√

2 i) = 0} / R[X], I6 = {f(X) ∈ R[X] : f(ζ3) = 0} / R[X].

1.12. Consider the inclusion inc : Q −→ C and its extension to ε√2 : Q[X] −→ C. Determinethe image ε√2Q[X] ⊆ C. What is ε−√2Q[X] ⊆ C? Find ker ε√2 /Q[X] and ker ε−√2 /Q[X];are these maximal ideals?

1.13. Let ω = (−1 +√

3i)/2 ∈ C. Consider the inclusion inc : Q −→ C and its extensionto εω : Q[X] −→ C. Determine the image εω Q[X] ⊆ C. Determine ker εω /Q[X] and decidewhether it is maximal. Find another evaluation homomorphism with the same kernel and image.

1.14. Consider the inclusion inc : Q −→ C and its extension to εα : Q[X] −→ C where α isone of the 4 complex roots of the polynomial f(X) = X4 − 2 ∈ Q[X]. Determine the imageεαQ[X] ⊆ C and the ideal ker εα /Q[X]; is the latter ideal maximal? What happens if α isreplaced by one of the other roots of f(X)?

Repeat this problem starting with the inclusion of the real numbers into the complex num-bers inc : R −→ C and εα : R[X] −→ C.

1.15. Prove the final part of Example 1.52 by showing that there is an isomorphism of groupsAff1(k) ∼= Aut|(k[X]).

1.16. Let k be any field. Consider the 6 automorphisms αj : k(X) −→ k(X) (j = 1, . . . , 6)defined by

α1(f(X)) = f(X), α2(f(X)) = f(1−X), α3(f(X)) = f(1/X),

α4(f(X)) = f((X − 1)/X), α5(f(X)) = f(1/(1−X)), α6(f(X)) = f(X/(X − 1)).

Show that the set consisting of these elements is a subgroup Γ| 6 Aut|(k(X)) isomorphic tothe symmetric group S3. When k = F2, show that Γ| ∼= GL2(k).

1.17. Find the cyclotomic polynomial Φ20(X) and determine its complex roots.

CHAPTER 2

Fields and their extensions

2.1. Fields and subfields

Definition 2.1. Let K and L be fields and suppose that K ⊆ L is a subring. Then we saythat K is a subfield of L; L is also said to be an extension (field) of K. We write K 6 L toindicate this and K < L if K is a proper subfield of L, i.e., if K 6= L. We will also sometimeswrite L/K when discussing the extension of K to L.

An important fact about an extension of fields L/K is that L is a K-vector space whoseaddition is the addition in the field L while scalar multiplication is defined by

u · x = ux (u ∈ K, x ∈ L).

Definition 2.2. We will call dimK L the degree or index of the extension L/K and use thenotation [L : K] = dimK L. An extension of fields L/K is finite if [L : K] < ∞, otherwise it isinfinite.

Example 2.3. The extension C/R is finite, while R/Q and C/Q are infinite.

Proof. We haveC = {x + yi : x, y ∈ R},

so 1, i span C as a vector space over R; as i /∈ R, these elements are also linearly independentover R they form a basis, hence [C : R] = 2. The infiniteness of R/Q and C/Q are consequencesof the fact that any finite dimensional vector space over Q is countable, however R and C areuncountable. A basis for the Q-vector space R is known as a Hamel basis. ¤

Example 2.4. Consider the extension Q(√

2)/Q where

Q(√

2) = {x + y√

2 : x, y ∈ Q}.Then [Q(

√2) : Q] = 2.

Proof. The elements 1,√

2 clearly span the Q-vector space L. Now recall that√

2 /∈ Q. Ifthe elements 1,

√2 were linearly dependent we would have u + v

√2 = 0 for some u, v ∈ Q not

both zero; in fact it is easy to see that we must have u, v both non-zero. Thus we have√

2 = −u

v∈ Q,

which is false. Hence 1,√

2 are linearly independent and so form a basis for Q(√

2) over Q and[Q(

√2) : Q] = 2. ¤

If we have two extensions L/K and M/L then it is a straightforward to verify that K 6 M

and so we have another extension M/K.

Definition 2.5. Given two extensions L/K and M/L, we say that L/K is a subextensionof M/K and sometimes write L/K 6 M/L.

Theorem 2.6. Let L/K be a subextension of M/K.

i) If one or both of the dimensions [L : K] or [M : L] is infinite then so is [M : K].

21

22 2. FIELDS AND THEIR EXTENSIONS

ii) If the dimensions [L : K] and [M : L] are both finite then so is [M : K] and

[M : K] = [M : L] [L : K].

Proof. (i) If [M : K] is finite, choose a basis m1, . . . , mr of M over K. Now any elementu ∈ M can be expressed as

u = t1m1 + · · ·+ trmr,

where t1, . . . , tr ∈ K; but since K ⊆ L, this means that m1, . . . ,mr spans M over L and so[M : L] < ∞. Also L is a K-vector subspace of the finite dimensional K-vector space M , hence[L : K] < ∞.(ii) Setting r = [L : K] and s = [M : L], choose a basis `1, . . . `r of L over K and a basism1, . . . , ms of M over L.

Now let v ∈ M . Then there are elements y1, . . . , ys ∈ L for which

v = y1m1 + · · ·+ ysms.

But each yj can be expressed in the form

yj = x1j`1 + · · ·+ xrj`r

for suitable xij ∈ K. Hence,

v =s∑

j=1

(r∑

i=1

xijì

)mj =

s∑

j=1

r∑

i=1

xij(ìmj),

where each coefficient xij is in K. Thus the elements ìmj (i = 1, . . . , r, j = 1, . . . , s) span theK-vector space M .

Now suppose that for some tij ∈ K we haves∑

j=1

r∑

i=1

tij(ìmj) = 0.

On collecting terms we obtains∑

j=1

(r∑

i=1

tijì

)mj = 0,

where each coefficient∑r

i=1 tijì is in L. By the linear independence of the mj over L, thismeans that for each j,

r∑

i=1

tijì = 0.

By the linear independence of the ì over K, each tij = 0.Hence the ìmj form a basis of M over K and so

[M : K] = rs = [M : L] [L : K]. ¤

We will often indicate subextensions in diagrammatic form where larger fields always goabove smaller ones and the information on the lines indicates dimensions

M

[M :L]

[M :K]=[M :L] [L:K]

.)$Â

½¸

³

L

[L:K]

K

We often suppress ‘composite’ lines such as the dashed one. Such towers of extensions are ourmain objects of study. We can build up sequences of extensions and form towers of arbitrary

2.2. SIMPLE AND FINITELY GENERATED EXTENSIONS 23

length. Thus, if L1/K, L2/L1, . . . , Lk/Lk−1 is a such a sequence of extensions, there is adiagram

Lk

Lk−1

L1

K

2.2. Simple and finitely generated extensions

Definition 2.7. Let F be a field and K 6 F . Given elements u1, . . . , ur ∈ F we set

K(u1, . . . , ur) =⋂

K6L6Fu1,...,ur∈L

L

which is the smallest subfield in F that contains K and the elements u1, . . . , ur. The ex-tension K(u1, . . . , ur)/K is said to be generated by the elements u1, . . . , ur; we also say thatK(u1, . . . , ur)/K is a finitely generated extension of K. An extension of the form K(u)/K iscalled a simple extension of K with generator u.

We can extend this to the case of an infinite sequence u1, . . . , ur, . . . in F and denote byK(u1, . . . , ur, . . .) 6 F the smallest extension field of K containing all the elements ur.

It is not difficult to show that

(2.1) K(u1, . . . , ur) ={

f(u1, . . . , ur)g(u1, . . . , ur)

∈ F : f(X1, . . . , Xr), g(X1, . . . , Xr) ∈ K[X1, . . . , Xr], g(u1, . . . , ur) 6= 0}

.

Reordering the ui does not change K(u1, . . . , un).

Proposition 2.8. Let K(u)/K and K(u, v)/K(u) be simple extensions. Then

K(u, v) = K(u)(v) = K(v)(u).

More generally,K(u1, . . . , un) = K(u1, . . . , un−1)(un)

and this is independent of the order of the sequence u1, . . . , un.

Theorem 2.9. For a simple extension K(u)/K exactly one of the following conditions holds.

i) The evaluation at u homomorphism εu : K[X] −→ K(u) is a monomorphism and onpassing to the fraction field gives an isomorphism (εu)∗ : K(X) −→ K(u). In this case,K(u)/K is infinite and u is said to be transcendental over K.

ii) The evaluation at u homomorphism εu : K[X] −→ K(u) has a non-trivial kernelker εu = (p(X)) where p(X) ∈ K[X] is an irreducible monic polynomial of positive de-gree and the quotient homomorphism εu : K[X]/(p(X)) −→ K(u) is an isomorphism.In this case K(u)/K is finite with [K(u) : K] = deg p(X) and u is said to be algebraicover K.


Proof. If ker εu = (0) then all that needs checking is that (εu)∗ is an epimorphism; but asu is in the image of (εu)∗ this is obvious.

When ker εu 6= (0), Theorem 1.27(iv) implies that the image of εu is a subfield of K(u) andsince it contains u it must equal K(u). Hence εu is an isomorphism. Using Long Division, wefind that every element of K[X]/(p(X)) can be uniquely expressed as a coset of the form

f(X) + (p(X))

where deg f(X) < deg p(X). Hence every element of K[X]/(p(X)) can be uniquely expressedas a linear combination over K of the d cosets

1 + (p(X)), X + (p(X)), X2 + (p(X)), . . . , Xd−1 + (p(X)),

where d = deg p(X). Via the isomorphism εu under which εu(Xk + (p(X))) = uk, we see thatthe elements 1, u, . . . , ud−1 form a basis for K(u) over K. ¤

Example 2.10. Consider the extension Q(√

2,√

3)/Q. Then [Q(√

2,√

3) : Q] = 4.

Proof. By Example 2.4 we know that [Q(√

2) : Q] = 2. We have the following tower ofextensions.

Q(√

2,√

3)

[Q(√

2,√

3):Q(√

2)]

[Q(√

2,√

3):Q]=2[Q(√

2,√

3):Q(√

2)]Q(√

2)

2

Q

We will show that [Q(√

2,√

3) : Q(√

2)] = 2.Notice that if u ∈ Q(

√2,√

3) = Q(√

2)(√

3) then u = a + b√

3 for some a, b ∈ Q(√

2),so 1,

√3 span Q(

√2,√

3) over Q(√

2). But if these are linearly dependent then√

3 ∈ Q(√

2).Writing √

3 = v + w√

2

with v, w ∈ Q, we find that

v2 + 3w2 + 2vw√

2 = 3 ∈ Q,

and hence 2vw√

2 ∈ Q. The possibilities v = 0 or w = 0 are easily ruled out, while if v, w 6= 0then

√2 ∈ Q which is false. So 1,

√3 are linearly independent over Q(

√2) and therefore form a

basis of Q(√

2,√

3). This shows that [Q(√

2,√

3) : Q(√

2)] = 2, hence [Q(√

2,√

3) : Q] = 4. ¤

Remark 2.11. There are some other subfields ofQ(√

2,√

3) which are conveniently displayedin the following diagram.

Q(√

2,√

3)2

rrrrrrrrrr2

2

LLLLLLLLLL

Q(√

2)

2LLLLLLLLLLLLQ(√

3)

2

Q(√

6)

2rrrrrrrrrrrr

Q

One idea in the verification of Example 2.10 can be extended to provide a useful generalresult whose proof is left as an exercise.

2.2. SIMPLE AND FINITELY GENERATED EXTENSIONS 25

Proposition 2.12. Let p1, . . . , pn be a sequence of distinct primes pi > 0. Then√

pn /∈ Q(√

p1, . . . ,√

pn−1).

Hence [Q(√

p1, . . . ,√

pn) : Q(√

p1, . . . ,√

pn−1)] = 2 and [Q(√

p1, . . . ,√

pn) : Q] = 2n.

Example 2.13. For the extension Q(√

2, i)/Q we have [Q(√

2, i) : Q] = 4.

Proof. We know that [Q(√

2) : Q] = 2. Also, i /∈ Q(√

2) since i is not real and Q(√

2) 6 R.Since i2 + 1 = 0, we have Q(

√2, i) = Q(

√2)(i) and [Q(

√2, i) : Q(

√2)] = 2. Using the formula

[Q(√

2, i) : Q] = [Q(√

2, i) : Q(√

2)] [Q(√

2) : Q],

we obtain [Q(√

2, i) : Q] = 4. ¤

This example also has several other subfields, with only Q(√

2) = Q(√

2, i) ∩ R being asubfield of R.

C2

∞R

∞ Q(√

2, i)2

ttttttttt2

2

KKKKKKKKKK

Q(√

2)

2 KKKKKKKKKKKQ(i)

2

Q(√

2 i)

2rrrrrrrrrrr

Q



2.1. Let p ∈ N be an prime. Show that the extension Q(√

p)/Q has [Q(√

p) : Q] = 2.

2.2. Let p, q > 0 be distinct primes. Show that [Q(√

p,√

q) : Q(√

p)] = 2.

2.3. Prove Proposition 2.12 by induction on n.

2.4. Let K a field with charK 6= 2 and suppose that L/K is an extension. If a, b ∈ K aredistinct, suppose that u, v ∈ L satisfy u2 = a and v2 = b. Show that K(u, v) = K(u + v).[Hint: show that u± v 6= 0 and deduce that u− v ∈ K(u + v); then show that u, v ∈ K(u + v).]

2.5. Show that [Q(i) : Q] = 2.

2.6. Show that [Q(√

3, i) : Q] = 4. Find the three subfields L 6 Q(√

3, i) with [L : Q] = 2 anddisplay their relationship in a diagram, indicating which ones are subfields of R.

2.7. Let ζ5 = e2πi/5 ∈ C.(a) Explain why [Q(ζ5) : Q] = 4.(b) Show that cos(2π/5), sin(2π/5) i ∈ Q(ζ5).(c) Show that for t ∈ R,

cos 5t = 16 cos5 t− 20 cos3 t + 5 cos t.

(d) Show that the numbers cos(2kπ/5) with k = 0, 1, 2, 3, 4 are roots of the polynomial

f(X) = 16X5 − 20X3 + 5X − 1 = (X − 1)(4X2 + 2X − 1)2

and deduce that [Q(cos(2π/5)) : Q] = 2.(e) Display the relationship between the fields Q, Q(cos(2π/5)), and Q(ζ5) in a suitable

diagram.

2.8. This question is for those who like lots of calculation (or using Maple). Let ζ7 = e2πi/7 ∈ C.(a) Explain why [Q(ζ7) : Q] = 6.(b) Show that cos(2π/7), sin(2π/7) i ∈ Q(ζ7).(c) Show

cos 7t = 64 cos7 t− 112 cos5 t + 56 cos3 t− 7 cos t.

Show that the numbers cos(2kπ/7) with k = 0, 1, . . . , 6 are roots of the polynomial

f(X) = 64X7 − 112X5 + 56X3 − 7X − 1 = (X − 1)(8X3 + 4X2 − 4X − 1)2

and deduce that [Q(cos(2π/7)) : Q] = 3.(d) Show that sin(2π/7) i is a root of

g(X) = 7X + 56X3 + 112X5 + 64X7 = X(7 + 56X2 + 112X4 + 64X6).

and that 7+56X2 +112X4 +64X6 ∈ Q[X] is irreducible. What is [Q(sin(2π/7) i) : Q]?(e) Display the relationship between the fields Q, Q(cos(2π/7)), Q(sin(2π/7) i) and Q(ζ7)

in a diagram.(f) Is i ∈ Q(ζ7)?

CHAPTER 3

Algebraic extensions of fields

3.1. Algebraic extensions

Let L/K be an extension of fields. From Theorem 2.9(ii), recall the following notion.

Definition 3.1. An element t ∈ L is algebraic over K if there is a non-zero polynomialp(X) ∈ K[X] for which p(t) = 0.

Notice in particular that for an element t ∈ K, the polynomial p(X) = X − t ∈ K[X]satisfies p(t) = 0, so such a t is algebraic over K.

Theorem 2.9 allows us to characterize algebraic elements in other ways.

Proposition 3.2. Let t ∈ L. Then the following conditions are equivalent.

i) t is algebraic over K.ii) The evaluation homomorphism εt : K[X] −→ L has non-trivial kernel.iii) The extension K(t)/K is finite dimensional.

Definition 3.3. If t ∈ L is algebraic over K then by Proposition 3.2,

ker εt = (minpolyK,t(X)) 6= (0),

where minpolyK,t(X) ∈ K[X] is a monic polynomial called the minimal polynomial of t overK. The degree of minpolyK,t(X) is called the degree of t over K and denoted degK t.

Proposition 3.4. If t ∈ L is algebraic over K then

[K(t) : K] = deg minpolyK,t(X) = degK t.

Proof. This follows from Theorem 2.9(ii). ¤

Remark 3.5. Suppose that t ∈ L is algebraic over K and p(X) ∈ ker εt with deg p(X) =deg minpolyK,t(X). Then minpolyK,t(X) | p(X) and we must have

p(X) = uminpolyK,t(X)

for some u ∈ K. In particular, when p(X) is monic,

p(X) = minpolyK,t(X).

We will often use this without further comment.

Example 3.6. Consider C/Q. The minimal polynomial of√

2 ∈ C over Q is

minpolyQ,√

2(X) = X2 − 2.

Proof. Clearly X2 − 2 ∈ ker ε√2 since (√

2)2 − 2 = 0. By Example 2.4,

deg minpolyQ,√

2(X) = [Q(√

2) : Q] = 2,

hence minpolyQ,√

2(X) = X2 − 2. ¤

Example 3.7. Consider C/Q. The minimal polynomial of i ∈ C over Q is X2 + 1.

27

28 3. ALGEBRAIC EXTENSIONS OF FIELDS

Proof. Clearly X2 + 1 ∈ ker εi since i2 + 1 = 0. As [Q(i) : Q] = 2, we have

minpolyQ,i(X) = X2 + 1. ¤

Example 3.8. Consider C/Q. Find the minimal polynomial of the primitive 6-th root ofunity, ζ6 ∈ C over Q.

Solution. Recall from Example 1.39 that ζ6 is a root of the irreducible cyclotomic poly-nomial

Φ6(X) = X2 −X + 1.

Then Φ6(X) ∈ ker εζ6 so minpolyQ,ζ6(X) | Φ6(X). Since Φ6(X) is irreducible and monic, wemust have

minpolyQ,ζ6(X) = Φ6(X)

and so degQ ζ6 = 2. ¤

Example 3.9. Consider C/Q. Find the minimal polynomial of√

2 +√

3 over Q.

Solution. Notice that√

3−√

2 =(√

3−√2)(√

3 +√

2)(√

3 +√

2)=

1√2 +

√3∈ Q(

√2 +

√3).

So we have√

2 =12

((√

2 +√

3)− (√

3−√

2)∈ Q(

√2 +

√3),

√3 =

12

((√

2 +√

3) + (√

3−√

2)∈ Q(

√2 +

√3),

hence Q(√

2,√

3) 6 Q(√

2 +√

3). Since Q(√

2 +√

3) 6 Q(√

2,√

3) we must have

Q(√

2 +√

3) = Q(√

2,√

3).

Referring to Example 2.10 we see that

degQ(√

2 +√

3) = 4.

Let us find a non-zero polynomial in ker ε√2+√

3 /Q[X].Referring to Example 2.10 or Proposition 2.12 we see that

√2 +

√3 /∈ Q(

√2), hence

degQ(√

2)(√

2 +√

3) = 2.

One polynomial in ker ε√2+√

3 /Q(√

2)[X] is

(X − (√

2 +√

3))(X − (√

2−√

3)) = X2 − 2√

2X − 1.

Since this is monic and of degree 2,

minpolyQ(√

2),√

2+√

3(X) = X2 − 2√

2X − 1.

Similarly,minpolyQ(

√2),−√2+

√3(X) = X2 + 2

√2X − 1.

Consider

p(X) = minpolyQ(√

2),√

2+√

3(X)minpolyQ(√

2),−√2+√

3(X)

= (X2 − 2√

2X − 1)(X2 + 2√

2X − 1)

= X4 − 10X2 + 1.

Then p(√

2 +√

3) = 0 so p(X) ∈ ker εt. Since deg p(X) = 4 and p(X) is monic, we have

minpolyQ,√

2+√

3(X) = X4 − 10X2 + 1. ¤

3.1. ALGEBRAIC EXTENSIONS 29

Definition 3.10. Let L/K be a finite extension. An element u ∈ L for which L = K(u) iscalled a primitive element for the extension L/K.

Later we will see that if charK = 0 then all finite extensions L/K have primitive elements.

Lemma 3.11. Let L/K be a finite extension and u ∈ L. Then u is a primitive element forL/K if and only if degK u = [L : K].

Proof. K(u) ⊆ L is a finite dimensional K-vector subspace. Then K(u) = L if and onlydimK K(u) = dimK L. Since degK u = dimK K(u) and [L : K] = dimK L the result follows. ¤

In all the examples we have seen so far the minimal polynomials have actually been irre-ducible. This is no accident.

Proposition 3.12. Let t ∈ L be algebraic over K. Then minpolyK,t(X) ∈ K[X] is irre-ducible.

Proof. Since K[X]/(minpolyK,t(X)) ∼= K(t) is a field, minpolyK,t(X) is irreducible byTheorem 1.27(iv). ¤

The following result will often be useful and can be proved by induction on n.

Lemma 3.13. Let L/K be an extension and suppose that u1, . . . , un ∈ L are algebraic. ThenK(u1, . . . , un)/K is a finite extension.

We now come to an important notion for extensions.

Definition 3.14. The extension L/K is algebraic or L is algebraic over K if every elementt ∈ L is algebraic over K.

Proposition 3.15. Let L/K be a finite extension. Then L/K is algebraic.

Proof. Let t ∈ L. Since the K-vector space L is finite dimensional the powers 1, t, . . . , tn, . . .

must be linearly dependent over K, hence for suitable coefficients cj ∈ K not all zero and somem > 1 we have

c0 + c1t + · · ·+ cmtm = 0.

But this means that t is algebraic over K. ¤

Proposition 3.16. Let M/L and L/K be algebraic extensions. Then the extension M/K

is algebraic.

Proof. Let u ∈ M . Then u is algebraic over L, so there is a polynomial

p(X) = p0 + p1X + · · ·+ pmXm ∈ L[X]

of positive degree with p(u) = 0. By Lemma 3.13, the extension K(p0, . . . , pm)/K is finite andso is K(p0, . . . , pm, u)/K(p0, . . . , pm). By Theorem 2.6(ii), K(p0, . . . , pm, u)/K is finite, so byProposition 3.15, u is algebraic over K. ¤

Definition 3.17. For an extension L/K, let

Lalg = {t ∈ L : t is algebraic over K} ⊆ L.

Proposition 3.18. For an extension L/K, Lalg is a subfield containing K and Lalg/K isalgebraic.

Proof. Clearly K ⊆ Lalg. We must show that Lalg 6 L.Let u, v ∈ Lalg. Then by Lemma 3.13, K(u, v)/K is a finite dimensional extension, hence

every element of K(u, v) is algebraic over K. In particular, u + v and uv are in K(u, v) and ifu 6= 0, u−1 is also in K(u, v). Therefore u + v, uv and u−1 are all algebraic over K. ¤


Example 3.19. In the extension C/Q we can consider Calg 6 C which is called the subfieldof algebraic numbers. Similarly, in the extension R/Q the subfield

Ralg = Calg ∩ R 6 C

consists of all the real algebraic numbers. Elements of C−Calg are called transcendental complexnumbers; examples are e and π. The sets Calg and Ralg are both countable, whereas C and Rare uncountable, so there are in fact many more transcendental numbers but it can be hard todetermine whether a given number is transcendental or not. A more usual notation for Calg

is Q since this is the algebraic closure of Q which will be discussed later. When dealing withalgebraic extensions of Q we will usually work with subfields of Q = Calg.

We end this section with a technical result.

Proposition 3.20. Let L/K be a finite simple extension, L = K(u). Then there are onlyfinitely many subextensions F/K 6 L/K.

Proof. Consider the minimal polynomial minpolyK,u(X) ∈ K[X]. Now for any subexten-sion F/K 6 L/K we can also consider

minpolyF,u(X) = c0 + c1X + · · ·+ ck−1Xk−1 + Xk ∈ F [X],

which divides minpolyK,u(X) in F [X]. The Unique Factorization Property 1.29 implies thatminpolyK,u(X) has only finitely many monic divisors in L[X], so there are only a finite numberof possibilities for minpolyF,u(X). Now let F0 = K(c0, c1, . . . , ck−1) be the extension field of K

generated by the coefficients of minpolyF,u(X). Then F0 6 F and so minpolyF,u(X) ∈ F0[X]is irreducible since it is irreducible in F [X]; hence minpolyF,u(X) = minpolyF0,u(X). As L =F (u), we have

[L : F ] = deg minpolyF,u(X) = deg minpolyF0,u(X) = [L : F0].

Hence F = F0.This shows that there are only finitely many subextensions F/K 6 L/K, each of which has

the form K(a0, a1, . . . , a`−1), where

a0 + a1X + · · ·+ a`−1X`−1 + X` ∈ L[X]

is a factor of minpolyK,u(X) in L[X]. ¤

3.2. Kronecker’s Theorem

We can now answer a basic question. Let K be a field and p(X) ∈ K[X] be a polynomialof positive degree.

Question 3.21. Is there an extension field L/K for which p(X) has a root in L?

A stronger version of this question is the following.

Question 3.22. Is there an extension field E/K for which p(X) factorizes into linear factorsin E[X]?

Definition 3.23. We say that p(X) splits in E/K if it factorizes into linear factors in E[X].

Of course, if we have such a field E then the distinct roots u1, . . . , uk of p(X) in E generatea subfield K(u1, . . . , uk) 6 E which is the smallest subfield of E that answers Question 3.22.

Definition 3.24. Such a minimal extension of K is called a splitting field of p(X) over K

and we will sometimes denote it by K(p(X)) or Kf .

We already know how to answer Question 3.21.

3.2. KRONECKER’S THEOREM 31

Theorem 3.25 (Kronecker’s Theorem: first version). Let K be a field and p(X) ∈ K[X] bea polynomial of positive degree. Then there is a finite extension L/K for which p(X) has a rootin L.

Proof. We begin by factorizing p(X) ∈ K[X] into irreducible monic factors qj(X) togetherwith a constant factor c:

p(X) = cq1(X) · · · qr(X).

Now for any j we can form the quotient field K[x]/(qj(X)) which is finite dimensional (simple)extension of K and in which the coset X + (qj(X)) satisfies the equation

qj(X + (qj(X))) = 0 + (qj(X)).

Hence p(X) has a root in K[x]/(qj(X)).Of course, this construction is only interesting if qj(X) to has degree bigger than 1 since a

linear polynomial already has a root in K. ¤

To answer Question 3.22 we iterate this construction. Namely, having found one root u1 inan extension L1/K we discard the linear factor X − u1 and consider the polynomial

p1(X) = p(X)/(X − u1) ∈ L1[X].

We can repeat the argument to form a finite extension of L1 (and hence of K) containing aroot of q1(X) and so on. At each stage we either already have another root in L1 or we need toenlarge the field to obtain one.

Theorem 3.26 (Kronecker’s Theorem: second version). Let K be a field and p(X) ∈ K[X]be a polynomial of positive degree. Then there is a finite extension E/K which is a splittingfield of p(X) over K.

In practise we often have extension fields ‘lying around in nature’ containing roots and wecan work inside of these. When working over Q (or any other subfield of C) we can always findroots in C by the Fundamental Theorem of Algebra. We then refer to a subfield of C which isa splitting field as the splitting subfield.

Example 3.27. Find a splitting field E/Q for p(X) = X4−4 over Q and determine [E : Q].

Solution. Notice thatp(X) = (X2 − 2)(X2 + 2),

so we first adjoin the roots ±√2 of (X2 − 2) to form Q(√

2,−√2) = Q(√

2) which gives anextension Q(

√2)/Q of degree 2.

Next consider the polynomial X2 + 2 ∈ Q(√

2)[X]. The complex roots of X2 + 2 are ±√2i

and these are not real, so this polynomial is irreducible in Q[X]. Hence we need to considerQ(√

2,√

2i) = Q(√

2, i) and the extension Q(√

2, i)/Q(√

2) which has degree 2.

C

∞

Q(√

2, i)

adjoin roots of X2 + 2 2

Q(√

2)

adjoin roots of X2 − 2 2

Q

Thus the splitting subfield of p(X) over Q in C is Q(√

2, i) and [Q(√

2, i) : Q] = 4. ¤


Of course we could have started by adjoining roots of X2 + 2 and then of X2 − 2, giving atower

C

∞

Q(√

2, i)

adjoin roots of X2 − 2 2

Q(√

2i)

adjoin roots of X2 + 2 2

Q

An important point is that if a splitting field exists inside of a given extension field F/K, it isunique.

Proposition 3.28. Let F/K be an extension field and p(X) ∈ K[X]. If E1, E2 6 F aresplitting subfields for p(X) over K then E1 = E2.

Proof. Let u1, . . . , uk ∈ F be the distinct roots of p(X) in F . By definition, K(u1, . . . , uk)is the smallest subfield containing K and all the uj . But K(u1, . . . , uk) must be contained inany splitting subfield, so E1 = K(u1, . . . , uk) = E2. ¤

Since we frequently encounter quadratic polynomials we record a useful result on roots ofsuch polynomials. Recall that p(X) = aX2 + bX + c ∈ K[X] is quadratic if a 6= 0 and itsdiscriminant is

∆ = b2 − 4ac ∈ K.

The proof of the next result is the standard one which works provided 2 has an inverse in K,i.e., when charK 6= 2.

Proposition 3.29. Let K be a field of characteristic different from 2. Then the quadraticpolynomial p(X) = aX2 + bX + c ∈ K[X] has

• no roots in K if ∆ is not a square in K;• one root −b/2 = −2−1b if ∆ = 0;• two distinct roots

−b + δ

2= 2−1(−b + δ),

−b− δ

2= 2−1(−b− δ),

if ∆ = δ2 for some non-zero δ ∈ K.

In particular, the splitting field of p(X) over K is K if ∆ is a square in K and and K(δ)otherwise where δ ∈ K is one of the two square roots of ∆ in the algebraic closure K.

Example 3.30. Find a splitting field E/Q for p(X) = X3−2 over Q and determine [E : Q].

Solution. By the Eisenstein Irreducibility Criterion 1.34, p(X) is irreducible over Q. Oneroot of p(X) is 3

√2 ∈ R so we adjoin this to Q to form an extension Q( 3

√2)/Q of degree 3. Now

p(X) = (X − 3√

2)(X2 + 3√

2X + ( 3√

2)2)

and the second factor has the non-real complex roots 3√

2 ζ3 and 3√

2 ζ23 , which lie in the extension

Q( 3√

2, ζ3)/Q( 3√

2) of degree 2. So the splitting subfield of X3 − 2 in C over Q is Q( 3√

2, ζ3) for

3.3. MONOMORPHISMS BETWEEN EXTENSIONS 33

which [Q( 3√

2, ζ3) : Q] = 6.

C

∞

2

R

∞ Q( 3√

2, ζ3)

Q( 3√

2)

2iiiiiiiiiiiiiiiiiiiii

3

LLLLLLLLLLLLLLLLLLLLLLLLLLLLLQ( 3√

2 ζ3)

2qqqqqqqqqq

3

<<<<

<<<<

<<<<

<<<<

<<Q( 3√

2 ζ23 )

2

3 Q(ζ3)

3

888888888888888888

2

sssssssssss

Q

An alternative strategy would have been to adjoin one of the other roots 3√

2 ζ3 or 3√

2 ζ23

first. We could also have begun by adjoining ζ3 to form the extension Q(ζ3)/Q, but none ofthe roots of p(X) lie in this field so the extension Q( 3

√2, ζ3)/Q(ζ3) of degree 3 is obtained by

adjoining one and hence all of the roots. ¤

3.3. Monomorphisms between extensions

Definition 3.31. For extensions F/K and L/K, let MonoK(L,F ) denote the set of allmonomorphisms L −→ F which fix the elements of K.

Remark 3.32. We always have AutK(F ) ⊆ MonoK(F, F ) and MonoK(F, F ) is closed undercomposition but is not always a group since elements are not necessarily invertible. Of course,if F/K is finite, then MonoK(F, F ) = AutK(F ) since every injective K-linear transformation issurjective and so invertible.

We will also use the following notation.

Definition 3.33. Let F/K be an extension and p(X) ∈ K[X]. Set

Roots(p, F ) = {u ∈ F : p(u) = 0},the set of roots of p(X) in F . This is always a finite set which may of course be empty.

Suppose that p(X) ∈ K[X] is an irreducible polynomial which we might as well assume ismonic, and F/K an extension. Then if t ∈ F is a root of p(X) in the evaluation homomorphismεt : K[X] −→ F factors through the quotient monomorphism εt : K[X]/(p(X)) −→ F whoseimage is K(t) 6 F . Of course, there is one such monomorphism for each root of p(X) in F . Ifwe fix one such root t0 and identify K[X]/(p(X)) with K(t0) via εt0 , then each root of p(X) inF gives rise to a monomorphism εt ◦ ε−1

t0 : K(t0) −→ F .

K(t0)

eεt◦eε−1t0

**K[X]/(p(X))eεt0

∼=oo eεt // F

Notice that if ϕ : K[X]/(p(X)) −→ F is any homomorphism extending the identity functionon K, then the coset X + (p(X)) must be sent by ϕ to a root of p(X) in F , hence every suchhomomorphism arises this way. This discussion is summarized in the following result.


Proposition 3.34. Let K be p(X) ∈ K[X] be an irreducible polynomial and let t0 ∈ F bea root of p(X), where F/K is an extension. Then there is a bijection

Roots(p, F ) ←→ MonoK(K(t0), F )

given by t ←→ ϕt, where ϕt : K(t0) −→ F has the effect ϕt(t0) = t.

Example 3.35. Show that MonoQ(Q(√

2),C) has two elements.

Solution. We have Q(√

2) ∼= Q[X]/(X2−2) where X2−2 is irreducible over Q. Hence theQ-monomorphisms send

√2 to ±√2. In fact both possibilities occur, giving monomorphisms

id, α : Q(√

2) −→ C, whereα(a + b

√2) = a− b

√2.

We can replace C by Q(√

2) to obtain

MonoQ(Q(√

2),C) = MonoQ(Q(√

2),Q(√

2)) = AutQ(Q(√

2)).

We will see that this is not always true. ¤

Example 3.36. Show that MonoQ(Q( 3√

2),C) has 3 elements but MonoQ(Q( 3√

2),Q( 3√

2))only contains the identity function.

Solution. Here minpolyQ, 3√2(X) = X3 − 2 and there are 3 complex roots 3√

2, 3√

2 ζ3,3√

2 ζ23 . As two of these roots are not real, MonoQ(Q( 3

√2),Q( 3

√2)) contains only the identity

since Q( 3√

2) 6 R.Each of the above roots corresponds to one of the subfields Q( 3

√2), Q( 3

√2 ζ3) or Q( 3

√2 ζ2

3 )of C and there are 3 monomorphisms α0, α1, α2 : Q( 3

√2) −→ C given by

α0(a + b3√

2 + c( 3√

2)2) = a + b3√

2 + c( 3√

2)2,

α1(a + b3√

2 + c( 3√

2)2) = a + b3√

2 ζ3 + c( 3√

2)2 ζ23 ,

α1(a + b3√

2 + c( 3√

2)2) = a + b3√

2 ζ23 + c( 3

√2)2 ζ3.

The images of these mappings are α0Q( 3√

2) = Q( 3√

2), α1Q( 3√

2) = Q( 3√

2 ζ3), α2Q( 3√

2) =Q( 3√

2 ζ23 ). ¤

Proposition 3.37. Let F/K and L/K be extensions.

i) For p(X) ∈ K[X], each monomorphism α ∈ MonoK(L,F ) restricts to a functionαp : Roots(p, L) −→ Roots(p, F ) which is an injection.

ii) If α ∈ AutK(L,L) then αp : Roots(p, L) −→ Roots(p, L) is a bijection.

Proof. (i) For u ∈ Roots(p(X), L), we have

p(α(u)) = α(p(u)) = α(0) = 0,

so α maps Roots(p, L) into Roots(p, F ). Since α is an injection its restriction to Roots(p, L) ⊆ L

is also an injection.(ii) α and its inverse are both monomorphisms and induce functions αp : Roots(p, L) −→Roots(p, L) and (α−1)p : Roots(p, L) −→ Roots(p, L). It is straightforward to check that αp

and (α−1)p are inverse functions, hence they are bijections. ¤

Part (ii) says that any automorphism of L/K permutes the set of roots in L of a polynomialp(X) ∈ K[X]. This gives us a strong hold on the possible automorphisms. In the case of finite,or more generally algebraic, extensions it is the key to understanding the automorphism groupand this is a fundamental insight of Galois Theory.

Example 3.38. Determine MonoQ(Q( 3√

2, ζ3),C).

3.4. ALGEBRAIC CLOSURES 35

Solution. We have already met the extension Q( 3√

2, ζ3)/Q in Example 3.30 and we willmake use of information from there. We build up the list of monomorphisms in stages.

First consider monomorphisms that fix 3√

2 and hence fix the subfield Q( 3√

2). These formthe subset

MonoQ( 3√2)(Q( 3√

2, ζ3),C) ⊆ MonoQ(Q( 3√

2, ζ3),C).

We know that Q( 3√

2, ζ3) = Q( 3√

2)(ζ3) and that ζ3 is a root of the irreducible X2 + X + 1 ∈Q( 3√

2)[X]. So there are two monomorphisms id, α0 fixing Q( 3√

2), where α0 has the effect

α0 :(

3√

2 7−→ 3√

2ζ3 7−→ ζ2

3

).

Next we consider monomorphisms that send 3√

2 to 3√

2 ζ3. This time we have 2 distinct ways toextend to elements of MonoQ(Q( 3

√2, ζ3),Q( 3

√2, ζ3)) since again we can send ζ3 to either ζ3 or

ζ23 . The possibilities are

α1 :(

3√

2 7−→ 3√

2 ζ3

ζ3 7−→ ζ3

), α′1 :

(3√

2 7−→ 3√

2 ζ3

ζ3 7−→ ζ23

).

Finally we consider monomorphisms that send 3√

2 to 3√

2 ζ23 . There are again 2 possibilities

α2 :(

3√

2 7−→ 3√

2 ζ23

ζ3 7−→ ζ3

), α′2 :

(3√

2 7−→ 3√

2 ζ23

ζ3 7−→ ζ23

).

These are all 6 of the required monomorphisms. It is also the case here that

MonoQ(Q( 3√

2, ζ3),C) = MonoQ(Q( 3√

2, ζ3),Q( 3√

2, ζ3)) = AutQ(Q( 3√

2, ζ3)),

so these form a group. It is a nice exercise to show that AutQ(Q( 3√

2, ζ3)) ∼= S3, the symmetricgroup on 3 objects. It is also worth remarking that |AutQ(Q( 3

√2, ζ3))| = [Q( 3

√2, ζ3)) : Q]. ¤

We end this section with another useful result.

Proposition 3.39. Let L/K be an extension and α ∈ MonoK(L,L). Then α restricts toan automorphism αalg : Lalg −→ Lalg.

Proof. Suppose that u ∈ Lalg, say p(u) = 0 for some p(X) ∈ K[X] of positive degree.Then

p(α(u)) = α(p(u)) = α(0) = 0,

so α maps Lalg ⊆ L into itself and therefore gives rise to a restriction αalg : Lalg −→ Lalg whichis also a monomorphism. We must show that αalg is a bijection by showing it is surjective.

Let v ∈ Lalg and suppose that q(v) = 0 for some q(X) ∈ K[X] of positive degree. NowRoots(q, L) 6= ∅ since it contains v, and it is also finite. Then αq : Roots(q, L) −→ Roots(q, L)is a bijection by Proposition 3.37(ii), hence v = αq(w) = α(w) for some w ∈ Roots(q, L) ⊆ Lalg.This shows that v ∈ im α and so αalg is surjective. ¤

3.4. Algebraic closures

An important property of the complex numbers is that C is algebraically closed.

Theorem 3.40. Every non-constant polynomial p(X) ∈ C[X] has a root in C.

Corollary 3.41. Every non-constant polynomial p(X) ∈ C[X] has a factorization

p(X) = c(X − u1) · · · (X − ud),

where c, u1, . . . , ud ∈ C and this is unique apart from the order of the roots uj.

It is natural to pose the following question.

Question 3.42. Let K be a field. Is there an algebraically closed field F containing K?


By taking F alg we might as well ask that such a field be algebraic over K.

Definition 3.43. Let K be a field. An extension F/K is called an algebraic closure of K

if F is algebraic over K and algebraically closed.

Theorem 3.44. Let K be a field.

i) There is an algebraic closure of K.ii) Let F1 and F2 be algebraically closures of K. Then there is an isomorphism ϕ : F1 −→

F2 which fixes the elements of K.

K

~~}}}}

}}}}

ÃÃAAA

AAAA

A

F1ϕ // F2

Hence algebraic closures are essentially unique.

Proof. See [2] for a proof using Zorn’s Lemma 3.48 which is logically equivalent to theAxiom of Choice. ¤

Because of this uniqueness we usually fix on an algebraic closure of K and write K orKalg-cl. We already know the example C = C. There are some easy consequences of this. Wewill temporarily write E1

.= E2 to indicate that for extensions E1/K and E2/K there is anisomorphism E1 −→ E2 fixing the elements of K.

Proposition 3.45. Let K be a field.

i) If L/K is an algebraic extension, then L.= K.

ii) If L/K is an extension, then so is L/K and (L)alg .= K.

Proof. (i) By Proposition 3.16, every element of L is algebraic over K. Since L is alge-braically closed it is an algebraic closure of K.(ii) Every non-constant polynomial (L)alg has a root in (L) and by Proposition 3.16, all of itsroots are in fact algebraic over K since (L)alg is. Hence these roots lie in (L)alg, showing thatit is algebraically closed. ¤

For example, we have Q = Calg and R = C.There is a stronger result than Theorem 3.44(ii), the Monomorphism Extension Theorem,

which we will find useful. Again the proof uses Zorn’s Lemma which we will state. First weneed some definitions.

Definition 3.46. A partially ordered set (X,4) consists of a set X and a binary relation4 for which whenever x, y, z ∈ X,

• x 4 x;• if x 4 y and y 4 z then x 4 z;• if x 4 y and y 4 x then x = y.

(X,4) is totally ordered if for every pair x, y ∈ X, at least one of x 4 y or y 4 x is true.

Definition 3.47. Let (X, 4) be a partially ordered set and Y ⊆ X. An element y ∈ X

is an upper bound for Y if for every y ∈ Y , y 4 y. An upper bound for itself X is a maximalelement of X.

Theorem 3.48 (Zorn’s Lemma). Let (X, 4) be a partially ordered set in which every totallyordered subset has an upper bound. Then X has a maximal element.

3.4. ALGEBRAIC CLOSURES 37

Theorem 3.49 (Monomorphism Extension Theorem). Let M/K be an algebraic extensionand L/K 6 M/K. Suppose that ϕ0 : L −→ K is a monomorphism fixing the elements of K.Then there is an extension of ϕ0 to a monomorphism ϕ : M −→ K.

K

M

ϕ

88

L

ϕ0

@@¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢

K= // K

Proof. We consider the set X consisting of all pairs (F, θ), where F/L 6 M/L and θ : F −→K extends ϕ0. We order X using the relation 4 for which (F1, θ1) 4 (F2, θ2) whenever F1 6 F2

and θ2 extends θ1. Then (X, 4) is a partially ordered set.Suppose that Y ⊆ X is a totally ordered subset. Let

F =⋃

(F,θ)∈Y

F.

Then F /L 6 M/L. Also there is a function θ : F −→ K defined by

θ(u) = θ(u)

whenever u ∈ F for (F, θ) ∈ Y . It is routine to check that if u ∈ F ′ for (F ′, θ′) ∈ Y then

θ′(u) = θ(u),

so θ is well-defined. Then for every (F, θ) ∈ Y we have (F, θ) 4 (F , θ), so (F , θ) is an upperbound for Y . Now by Zorn’s Lemma there must be a maximal element of X, (M0, θ0).

Suppose that M0 6= M , so there is an element u ∈ M for which u /∈ M0. Since M is algebraicover K it is also algebraic over M0, hence u is algebraic over M0. If

minpolyM0,u(X) = a0 + · · ·+ an−1Xn−1 + Xn,

then the polynomial

f(X) = θ0(a0) + · · ·+ θ0(an−1)Xn−1 + Xn ∈ (θ0M0)[X]

is also irreducible and so it has a root v in K (which is also an algebraic closure of θ0M0 6K). The Homomorphism Extension Property 1.18 of the polynomial ring M0[X] applied tothe monomorphism θ0 : M0 −→ K yields a homomorphism θ′0 : M0[X] −→ K extending θ0

and having θ′0(u) = v. This factors through the quotient ring M0[X]/(minpolyM0,u(X)) togive a monomorphism θ′′0 : M0(u) −→ K extending θ0. But then (M0, θ0) 4 (M0(u), θ′′0) and(M0, θ0) 6= (M0(u), θ′′0), contradicting the maximality of (M0, θ0). Hence M0 = M and so wecan take ϕ = θ0. ¤

Example 3.50. Let u ∈ K be and suppose that p(X) = minpolyK,u(X) ∈ K[X]. Then forany other root of p(X), v ∈ K, there is a monomorphism ϕv : K(u) −→ K with ϕv(u) = v.This extends to a monomorphism ϕ : K −→ K.

Definition 3.51. Let u, v ∈ K. Then v is conjugate to u over K if there is a monomorphismϕ : K −→ K for which v = ϕ(u); we also say that v is a conjugate of u over K.

Lemma 3.52. If u, v ∈ K, then v is conjugate to u over K if and only if minpolyK,u(v) = 0.


Proof. Suppose that v = ϕ(u) for some ϕ ∈ MonoK(K, K). If

minpolyK,u(X) = a0 + a1X + · · ·+ ad−1Xd−1 + Xd,

thena0 + a1u + · · ·+ ad−1u

d−1 + ud = 0

and so

a0 + a1v + · · ·+ ad−1vd−1 + vd = ϕ(a0 + a1u + · · ·+ ad−1u

d−1 + ud) = 0.

The converse follows from Example 3.50. ¤

3.5. Multiplicity of roots and separability

Let K be a field. Suppose that f(X) ∈ K[X] and u ∈ K is a root of f(X), i.e., f(u) = 0.Then we can factor f(X) as f(X) = (X − u)f1(X) for some f1(X) ∈ K[X].

Definition 3.53. If f1(u) = 0 then u is a multiple or repeated root of f(X). If f1(u) 6= 0then u is a simple root of f(X).

We need to understand more clearly when an irreducible polynomial has a multiple rootsince this turns out to be important in what follows. Consider the formal derivative on K[X],i.e., the function ∂ : K[X] −→ K[X] given by

∂(f(X)) = f ′(X) = a1 + 2a2X + · · ·+ dadXd−1,

where f(X) = a0 + a1X + a2X2 + · · ·+ adX

d with aj ∈ K.

Proposition 3.54. The formal derivative ∂ : K[X] −→ K[X] has the following properties.

i) ∂ is K-linear.ii) ∂ is a derivation, i.e., for f(X), g(X) ∈ K[X],

∂(f(X)g(X)) = ∂(f(X))g(X) + f(X)∂(g(X)).

iii) If charK = 0, then ker ∂ = K and ∂ is surjective.iv) If charK = p > 0, then

ker ∂ = {h(Xp) : h(X) ∈ K[X]}and im ∂ is spanned by the monomials Xk with p - (k + 1).

Proof. (i) This is routine.(ii) By K-linearity, it suffices to verify this for the case where f(X) = Xr and g(X) = Xs withr, s > 0. But then

∂(Xr+s) = (r + s)Xr+s−1 = rXr−1Xs + sXrXs−1 = ∂(Xr)Xs + Xr∂(Xs).

(iii) If f(X) = a0 + a1X + a2X2 + · · ·+ adX

d then

∂(f(X)) = 0 ⇐⇒ a1 = 2a2 = · · · = dad = 0.

So ∂(f(X)) = 0 if and only if f(X) = a0 ∈ K. It is also clear that every polynomial g(X) ∈ K[X]has the form g(X) = ∂(f(X) where f(X) is an anti-derivative of g(X).(iv) For a monomial Xm, ∂(Xm) = mXm−1 and this is zero if and only if p | m. Using this wesee that

∂(a0 + a1X + a2X2 + · · ·+ adX

d) = 0 ⇐⇒ am = 0 whenever p - m.

Also, im ∂ is spanned by the monomials Xk for which ∂(Xk+1) 6= 0, which are the ones withp - (k + 1). ¤

We now apply the formal derivative to detect multiple roots.

3.5. MULTIPLICITY OF ROOTS AND SEPARABILITY 39

Proposition 3.55. Let f(X) ∈ K[x] have a root u ∈ L where L/K is an extension. Thenu is a multiple root of f(X) if and only if f(X) and f ′(X) have a common factor of positivedegree in K[X] which vanishes at u.

Proof. Working in L[X], let f(X) = (X − u)f1(X). Then

f ′(X) = f1(X) + (X − u)f ′1(X),

so f ′(u) = f1(u). Hence u is a multiple root if and only if f(X) and f ′(X) have a commonfactor in L[X] and hence in K[X] which vanishes at u. ¤

Corollary 3.56. If f(X) is irreducible in K[X] then a root u is a multiple root if and onlyif f ′(X) = 0. In particular, this can only happen if charK > 0.

Corollary 3.57. If charK = 0 and f(X) is irreducible in K[X], then every root of f(X)is simple.

Example 3.58. For n > 1, show that each of the roots of f(X) = Xn − 1 in C is simple.

Solution. We have f ′(X) = ∂(Xn − 1) = nXn−1, so if ζ is any root of f(X) then f ′(ζ) =nζn−1 6= 0. ¤

Example 3.59. Show that 2i is a multiple root of f(X) = X4 + 8X2 + 16.

Solution. We have f ′(X) = 4X3 + 16X. Using the Long Division and the EuclideanAlgorithm we find that gcd(f(X), f ′(X)) = X4 + 4, where 2i is also a root of X4 + 4. Hence 2iis a multiple root of f(X).

In fact, X4 + 8X2 + 16 = (X4 + 4)2, so this is obvious. ¤Example 3.60. Let p > 0 be a prime and suppose that L/Fp is an extension. Show each of

that the roots of f(X) = Xp − 1 in L is multiple.

Solution. We have f ′(X) = ∂(Xp − 1) = pXn−1 = 0, so if ζ is any root of f(X) thenf ′(ζ) = 0. Later we will see that 1 is the only root of Xp − 1. ¤

Definition 3.61. An irreducible polynomial p(X) ∈ K[X] is separable over K if every rootof p(X) in an extension L/K is simple. By Corollary 3.56, this is equivalent to requiring thatp′(X) 6= 0. If u ∈ L is a multiple root of p(X), then the multiplicity of u in p(X) is the maximumm such that p(X) = (X − u)mq(X) for some q(X) ∈ L[X].

Proposition 3.62. Let K be a field and let K be an algebraic closure. If the irreduciblepolynomial p(X) ∈ K[X] has distinct roots u1, . . . , uk ∈ K, then the multiplicities of the uj areequal. Hence in K[X],

p(X) = c(X − u1)m · · · (X − uk)m,

where c ∈ K and m > 1.

Proof. Let u ∈ K be a root of p(X) and suppose that it has multiplicity m, so we canwrite p(X) = (X − u)mp1(X) where p1(X) ∈ K(u)[X] and p1(u) 6= 0.

Now let v ∈ K be any other root of p(X). By Proposition 3.34, there is a monomorphismϕv : K(u) −→ K for which ϕv(u) = v. When p(X) is viewed as an element of K(u)[X], thecoefficients of p(X) are fixed by ϕv. Then

ϕv((X − u)mp1(X)) = (X − u)mp1(X),

so(X − v)mp1(X) = (X − u)mp1(X),

where p1(X) ∈ K[X] is obtained applying ϕv to the coefficients of p1(X). Now by Corollary 1.30,(X − v)m must divide p1(X) in K[X], and therefore the multiplicity of v must be at least m.Interchanging the roles of u and v we find that the multiplicities of u and v are in fact equal. ¤


Corollary 3.63. Let K be a field and let K be an algebraic closure. If the irreduciblepolynomial p(X) ∈ K[X] has distinct roots u1, . . . , uk ∈ K which are all simple then in K[X],

p(X) = c(X − u1) · · · (X − uk),

where c ∈ K and k = deg p(X).

Corollary 3.64. Let K be a field and let u ∈ K. Then the number of distinct conjugatesof u is

deg minpolyK,u(X)m

,

where m is the multiplicity of u in minpolyK,u(X).

Definition 3.65. An algebraic element u ∈ L in an extension L/K is separable if its minimalpolynomial minpolyK,u(X) ∈ K[X] is separable.

Definition 3.66. An algebraic extension L/K is called separable if every element of L isseparable over K.

Example 3.67. An algebraic extension L/K of a field of characteristic 0 is separable byCorollary 3.57.

Definition 3.68. Let L/K be a finite extension. The separable degree of L over K is

(L : K) = |MonoK(L,K)|.Lemma 3.69. For a finite simple extension K(u)/K,

(L : K) = |Roots(minpolyK,u,K)|.If K(u)/K is separable, then [L : K] = (L : K).

Proof. This follows from Proposition 3.34 applied to the case L = K. ¤

Any finite extension L/K can be built up from a succession of simple extensions

(3.1) K(u1)/K, K(u1, u2)/K(u1), · · · , L = K(u1, . . . , uk)/K(u1, . . . , uk−1).

So we can use the following to compute (L : K) = (K(u1, . . . , uk) : K).

Proposition 3.70. Let L/K and M/L be finite extensions. Then

(M : K) = (M : L)(L : K).

Proof. For α ∈ MonoK(M, K) let αL ∈ MonoK(L,K) be its restriction to L. By theMonomorphism Extension Theorem 3.49, each element of MonoK(L,K) extends to a monomor-phism M −→ K, so every element β ∈ MonoK(L,K) has the form β = αL for some α ∈MonoK(M, K); since (L : K) = |MonoK(L,K)|, we need to show that the number of such α isalways (M : L) = |MonoK(M, K)|.

So given β ∈ MonoK(L,K), choose any extension to a monomorphism β : K −→ K; byProposition 3.39, β is an automorphism. Of course, restricting to M 6 K we obtain a monomor-phism M −→ K. Now for any extension β′ : M −→ K of β we can form the compositionβ−1 ◦ β′ : M −→ K; notice that if u ∈ L, then

β−1 ◦ β′(u) = β−1(β(u)) = u,

hence β−1 ◦ β′ ∈ MonoL(M, K). Conversely, each γ ∈ MonoL(M, K) gives rise to a monomor-phism β ◦ γ : M −→ K which extends β. In effect, this shows that there is a bijection

{extensions of β to monomorphism a M −→ K

} ←→ MonoL(M, K),

so (M : L) = |MonoL(M, K)| agrees with the number of extensions of β to a monomorphismM −→ K. Therefore we have the desired formula (M : K) = (M : L)(L : K). ¤

3.5. MULTIPLICITY OF ROOTS AND SEPARABILITY 41

Corollary 3.71. Let L/K be a finite extension. Then (L : K) | [L : K].

Proof. If L/K is a simple extension then by Propositions 3.62 and 3.34 we know that thisis true. The general result follows by building up L/K as a sequence of simple extensions asin (3.1) and then using Theorem 2.6(ii) which gives

[L : K] = [K(u1) : K] [K(u1, u2) : K(u1)] · · · [K(u1, . . . , uk) : K(u1, . . . , uk−1)].

For each k, (K(u1, . . . , uk) : K(u1, . . . , uk−1)) divides [K(u1, . . . , uk) : K(u1, . . . , uk−1)], so thedesired result follows. ¤

Proposition 3.72. Let L/K be a finite extension. Then L/K is separable if and only if(L : K) = [L : K].

Proof. Suppose that L/K is separable. If K 6 E 6 L, then for any u ∈ L, u is alge-braic over E, and in the polynomial ring E[X] we have minpolyE,u(X) | minpolyK,u(X). AsminpolyK,u(X) is separable, so is minpolyE,u(X), and therefore L/E is separable. Clearly E/K

is also separable. We have (L : K) = (L : E) (E : K) and [L : K] = [L : E] [E : K], so toverify that (L : K) = [L : K] it suffices to show that (L : E) = [L : E] and (E : K) = [E : K].Expressing L/K in terms of a sequence of simple extensions as in (3.1), we find that

(L : K) = (K(u1) : K) · · · (L : K(u1, . . . , uk−1)),

and

[L : K] = [K(u1) : K] · · · [L : K(u1, . . . , uk−1].

We can apply Lemma 3.69 to each of these intermediate separable simple extensions to obtain(L : K) = [L : K].

For the converse, suppose that (L : K) = [L : K]. We must show that for each u ∈ L, u isseparable. For the extensions K(u)/K and L/K(u) we have (L : K) = (L : K(u)) (K(u) : K)and [L : K] = [L : K(u)] [K(u) : K]. By Corollary 3.71, there are some positive integers r, s forwhich [L : K(u)] = r(L : K(u)) and [K(u) : K] = s(K(u) : K). Hence

(L : K(u))(K(u) : K) = rs(L : K(u))(K(u) : K),

which can only happen if r = s = 1. This shows that (K(u) : K) = [K(u) : K], so u isseparable. ¤

Proposition 3.73. Let L/K and M/L be finite extensions. Then M/K is separable if andonly if L/K and M/L are separable.

Proof. If M/K is separable then [M : K] = (M : K) and so by Proposition 3.70,

[M : L][L : K] = (M : L)(L : K).

This can only happen if [M : L] = (M : L) and [L : K] = (L : K), since (M : L) 6 [M : L] and(L : K) 6 [L : K]. By Proposition 3.72 this implies that L/K and M/L are separable.

Conversely, if L/K and M/L are separable then [M : L] = (M : L) and [L : K] = (L : K),so

[M : K] = [M : L][L : K] = (M : L)(L : K) = (M : K).

Therefore M/K is separable. ¤


3.6. The Primitive Element Theorem

Definition 3.74. For a finite simple extension L/K, an element u ∈ L is called a primitiveelement for the extension if L = K(u).

Theorem 3.75 (Primitive Element Theorem). Let L/K be a finite separable extension.Then L has a primitive element.

Proof. The case where K is a finite field will be dealt with in Proposition 5.16. So we willassume that K is infinite.

Since L is built up from a sequence of simple extensions it suffices to consider the caseL = K(u, v). Let p(X), q(X) ∈ K[X] be the minimal polynomials of u and v over K. Supposethat the distinct roots of p(X) in K are u = u1, . . . , ur, while the distinct roots of q(X) arev = v1, . . . , vs. By the separability assumption, r = deg p(X) and s = deg q(X).

Since K is infinite, we can choose an element t ∈ K for which

t 6= u− ui

vj − v

whenever j 6= 1. Then taking w = u+ tv ∈ L, we find that w 6= ui + tvj whenever j 6= 1. Definethe polynomial (of degree r)

h(X) = p(w − tX) ∈ K(w)[X] ⊆ L[X].

Then h(v) = p(u) = 0, but h(vj) 6= p(ui) = 0 for any j 6= 1 by construction of t, so none of theother vj is a zero of h(X).

Now since the polynomials h(X), q(X) ∈ K(w)[X] have exactly one common root in K,namely v, by separability their greatest common divisor in K(w)[X] is a linear polynomial whichmust be X − v, hence v ∈ K(w) and so u = w − tv ∈ K(w). This shows that K(u, v) 6 K(w)and therefore K(w) = K(u, v). ¤

Corollary 3.76. Let L/K be a finite separable extension of a field of characteristic 0.Then L has a primitive element.

Proof. Since Q 6 K, K is infinite and by Example 3.67 L/K is separable. ¤

To find a primitive element we can always use the menu provided in the proof of Theo-rem 3.75, however a ‘try it and see’ approach will usually suffice.

Example 3.77. Find a primitive element for the extension Q(√

3, i)/Q.

Solution. Consider√

3+ i. Then working over the subfield Q(√

3) 6 Q(√

3, i) we find thati /∈ Q(

√3) 6 R and

(X − (√

3 + i))(X − (√

3− i)) = X2 − 2√

3X + 4 ∈ Q(√

3)[X],

henceX2 − 2

√3X + 4 = minpolyQ(

√3),√

3+i(X).

Now taking(X2 − 2

√3X + 4)(X2 + 2

√3X + 4) = X4 − 4X2 + 16 ∈ Q[X],

we see that minpolyQ,√

3+i(X) | (X4 − 4X2 + 16) in Q[X]. Notice that

(√

3 + i)−1 =(√

3− i)(√

3 + i)(√

3− i)=

(√

3− i)3 + 1

=14(√

3− i) ∈ Q(√

3 + i),

since (√

3 + i)−1 ∈ Q(√

3 + i). Hence√

3 =12((√

3 + i) + (√

3− i)), i =12((√

3 + i)− (√

3− i)),

3.7. NORMAL EXTENSIONS 43

are both in Q(√

3 + i), showing that Q(√

3, i) 6 Q(√

3 + i) and so Q(√

3, i) = Q(√

3 + i). Thuswe must have deg minpolyQ,

√3+i(X) = 4, and so minpolyQ,

√3+i(X) = X4 − 4X2 + 16. ¤

There is a general phenomenon illustrated by Example 3.77.

Proposition 3.78. Let u ∈ K be separable over K. Then

minpolyK,u(X) = (X − α1(u)) · · · (X − αd(u)),

where α1, . . . , αd are the elements of MonoK(K(u), K). In particular, the polynomial

(X − α1(u)) · · · (X − αd(u)) ∈ K[X]

is in K[X] and is irreducible therein.

Proof. Since K(u) is separable then by Lemma 3.52,

d = deg minpolyK,u(X) = [K(u) : K] = (K(u) : K). ¤

In Example 3.77 we have

[Q(√

3, i) : Q] = [Q(√

3, i) : Q(√

3)][Q(√

3) : Q] = 2 · 2 = 4.

There are four monomorphisms αk : Q(√

3, i) −→ Q(√

3, i) given by

α1 = id, α2 =(√

3 7−→ √3

i 7−→ −i

), α3 =

(√3 7−→ −√3

i 7−→ i

), α2 =

(√3 7−→ −√3

i 7−→ −i

).

Then

α2(√

3 + i) = (√

3− i), α3(√

3 + i) = (−√

3 + i), α2(√

3 + i) = (−√

3− i),

so(X −

√3− i)(X −

√3 + i)(X +

√3− i)(X +

√3 + i) = X4 − 4X2 + 16 ∈ Q[X].

Hence this polynomial is irreducible. Then [Q(√

3 + i) : Q] = 4 and Q(√

3 + i) = Q(√

3, i).

3.7. Normal extensions

Definition 3.79. A finite extension E/K is normal if ϕE = E for every ϕ ∈ MonoK(E,K).

Remark 3.80. If E/K is a normal extension then whenever an irreducible polynomialp(X) ∈ K[X] has a root in E then it splits in E since by Lemma 3.52 each pair of roots of p(X)is conjugate over K and one can be mapped to the other by a monomorphism K −→ K whichmust send E into itself.

Proposition 3.81. A finite extension E/K is normal if and only if it is a splitting fieldover K for some polynomial f(X) ∈ K[X].

Proof. Suppose that E/K is normal. Then there is a sequence of extensions

K 6 K(u1) 6 K(u1, u2) 6 · · · 6 K(u1, . . . , un) = E

Construct a polynomial by taking

f(X) = minpolyK,u1(X)minpolyK,u2

(X) · · ·minpolyK,un(X).

Then by Remark 3.80, f(X) splits in E. Also E is generated by some of the roots of f(X).Hence E is a splitting field for f(X) over K.

Now suppose that E is a splitting field for g(X) ∈ K[X], so that E = K(v1, . . . , vk), wherev1, . . . , vk are the distinct roots of g(X) in E. Now any monomorphism θ ∈ MonoK(E, K) mustmap these roots to θ(v1), . . . , θ(vk) which are also roots of g(X) and therefore lie in E (seeProposition 3.34). Since θ permutes the roots vj , we have

θE = θK(v1, . . . , vk) = K(θ(v1), . . . , θ(vk)) = K(v1, . . . , vk) = E. ¤


This result makes it easy to recognize a normal extension since it is sufficient to describe itas a splitting field for some polynomial over K. In Chapter 4 we will see that separable normalextensions play a central role in Galois Theory.

45


3.1. Prove Proposition 3.2.

3.2. Finding splitting subfields E 6 C over Q and determine [E : Q] for each of the followingpolynomials.

p1(X) = X4−X2+1, p2(X) = X6−2, p3(X) = X4+2, p4(X) = X4+5X3+10X2+10X+5.

[Hint: for p4(X), consider p4(Y − 1) ∈ Q[Y ].]

3.3. Prove that AutQ(Q( 3√

2, ζ3)) ∼= S3, the symmetric group on 3 elements, as claimed in thesolution of Example 3.38. [Hint: work out the effect of each automorphism on the three roots ofthe polynomial X3 − 2.]

3.4. Let k be a field of characteristic chark = p > 0 and k(T ) be the field of rational functionsin T over k. Show that the polynomial g(X) = Xp − T ∈ k(T )[X] is irreducible and has amultiple root in k(T ). How does g(X) factor in k(T )[X]?

3.5. Find primitive elements for the extensions Q(√

5,√

10)/Q, Q(√

2, i)/Q, Q(√

3, i)/Q,Q( 4√

3, i)/Q, in each case finding it minimal polynomial over Q. [Hint: look for elements ofhigh degree over Q, or use the method of proof of Theorem 3.75.]

3.6. Prove the following converse of Proposition 3.20:Let L/K be a finite extension. If there are only finitely many subextensions F/K 6 L/K, thenL/K is simple, i.e., L = K(w) for some w ∈ L.[Hint: First deal with the case where L = K(u, v), then use induction on n to prove the generalcase L = K(u1, . . . , un).]

3.7. Let K be a field. Show that every quadratic (i.e., of degree 2) extension E/K is normal.Is such an extension always separable?

3.8. Let f(X) ∈ Q[X] be an irreducible polynomial of odd degree greater than 1 and with onlyone real root u ∈ R. Show that Q(u)/Q is not normal.

CHAPTER 4

Galois extensions and the Galois Correspondence

In this Chapter we will study the structure of Galois extensions and their associated Galoisgroups, in particular we will explain how these are related through the Galois Correspondence.

Throughout the chapter, let K be a field.

4.1. Galois extensions

Definition 4.1. A finite extension E/K is a (finite) Galois extension if it is normal andseparable.

From Section 3.5 we know that for such a Galois extension E/K, [E : K] = (E : K) and alsoevery monomorphism ϕ ∈ MonoK(E,K) maps E into itself, hence restricts to an automorphismof E which will be denoted ϕ|E .

K

E

ϕ>>~~~~~~~

ϕ|E

∼= //___ E

K= // K

Also, by the Monomorphism Extension Theorem 3.49, every automorphism α ∈ AutK(E) ex-tends to a monomorphism E −→ K fixing elements of K. So there is a bijection

MonoK(E, K) ←→ AutK(E)

and we have

(4.1) |AutK(E)| = (E : K) = [E : K].

Definition 4.2. For a finite Galois extension E/K, the group Gal(E/K) = AutK(E) iscalled the Galois group of the extension or the Galois group of E over K.

Of course, Equation (4.1) gives

(4.2) |Gal(E/K)| = (E : K) = [E : K].

We can also reformulate the notion of conjugacy introduced in Definition 3.51.

Definition 4.3. Let E/K a finite Galois extension and u, v ∈ E. Then v is conjugate to u

if there is a ϕ ∈ Gal(E/K) for which v = ϕ(u); we also say that v is a conjugate of u.

It is easy to see that for u, v ∈ K, there is a finite Galois extension E/K in which v is aconjugate of u if and only v is a conjugate of u over K in the old sense.

47

48 4. GALOIS EXTENSIONS AND THE GALOIS CORRESPONDENCE

4.2. Working with Galois groups

Let E/K be a finite Galois extension. Then we know that E is a splitting field for somepolynomial over K. We also know that E is a simple extension of K and so it is a splittingfield for the minimal polynomial of a generating element which has degree equal to [E : K]. Itis often convenient to use these facts to interpret elements of the Galois group as permutationsof the roots of some polynomial which splits over E.

Example 4.4. Describe the Galois group Gal(Q(√

2,√

3)/Q as a subgroup of the group ofpermutations of the roots of (X2 − 2)(X2 − 3) ∈ Q[X].

Solution. We have

[Q(√

2,√

3) : Q] = [Q(√

2,√

3) : Q(√

2)] [Q(√

2) : Q] = 4,

and the following non-trivial elements of the Galois group together with the element identityα1 = id:

α2 =

√2 7−→ −√2

−√2 7−→ √2√

3 7−→ √3

−√3 7−→ −√3

, α3 =

√2 7−→ √

2−√2 7−→ −√2√

3 7−→ −√3−√3 7−→ √

3

, α4 =

√2 7−→ −√2

−√2 7−→ √2√

3 7−→ −√3−√3 7−→ √

3

.

Writing the roots in the list√

2,−√2,√

3,−√3 and numbering them 1 to 4 accordingly, we seethat these automorphisms correspond to the following permutations in S4 expressed in cyclenotation:

α2 ←→ (1 2), α3 ←→ (3 4), α4 ←→ (1 2)(3 4). ¤

Example 4.5. Using a primitive element u for the extension, describe the Galois groupGal(Q(

√2,√

3)/Q as a subgroup of the group of permutations of the roots of minpolyQ,u(X) ∈Q[X].

Solution. We have Q(√

2,√

3) = Q(√

2 +√

3) and the conjugates of u =√

2 +√

3 are±√2±√3. Listing these as

√2 +

√3,√

2−√

3, −√

2 +√

3, −√

2−√

3,

and after numbering them accordingly, we find the correspondences

α2 ←→ (1 3)(2 4), α3 ←→ (1 2)(3 4), α4 ←→ (1 4)(2 3). ¤

We next summarize the properties of Galois groups that can be deduced from what we haveestablished so far.

Recollection 4.6. Recall that an action of a group G on a set X is transitive if for everypair of elements x, y ∈ X, there is an element g ∈ G such that y = gx (so there is only oneorbit); the action is faithful or effective if for every h ∈ G, there is a element z ∈ X such thathz 6= z.

For an extension F/K and a polynomial f(X) ∈ K[X], recall that Roots(f, F ) denotes theset of roots of f(X) in F .

Theorem 4.7. Let E/K be a finite Galois extension. Suppose that E is the splitting fieldof a separable irreducible polynomial f(X) ∈ K[X] of degree n. Then the following are true.

i) Gal(E/K) acts transitively and faithfully on Roots(f,E).ii) Gal(E/K) can be identified with a subgroup of the group of permutations of Roots(f, E).

If we order the roots u1, . . . , un then Gal(E/K) can be identified with a subgroup of Sn.iii) |Gal(E/K)| divides n! and is divisible by n.

4.2. WORKING WITH GALOIS GROUPS 49

As we have seen in Examples 4.4 and 4.5, in practise it is often easier to use a not necessarilyirreducible polynomial to determine and work with a Galois group.

Example 4.8. The Galois extension Q(ζ8)/Q has degree [Q(ζ8) : Q] = 4 and it has thefollowing automorphisms apart from the identity:

α : ζ8 7−→ ζ38 , β : ζ8 7−→ ζ5

8 , γ : ζ8 7−→ ζ78 .

If we list the roots of the minimal polynomial

minpolyQ,ζ(X) = Φ8(X) = X4 + 1

in the order ζ8, ζ38 , ζ5

8 , ζ78 , we find that these automorphisms correspond to the following permu-

tations in S4:α ←→ (1 2)(3 4), β ←→ (1 3)(2 4), γ ←→ (1 4)(2 3).

So the Galois group Gal(Q(ζ8)/Q) corresponds to

{id, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} 6 S4.

Noticing that

ζ8 =1√2

+1√2i,

we easily find that√

2, i ∈ Q(ζ8); hence Q(√

2, i) 6 Q(ζ8). Since [Q(√

2, i) : Q] = 4, we haveQ(√

2, i) = Q(ζ8). Notice that Q(√

2, i) is the splitting field of f(X) = (X2 − 2)(X2 + 1) overQ. Now list the roots of f(X) in the order

√2,−√2, i,−i, and observe that

α :

√2 7−→ −√2

−√2 7−→ √2

i 7−→ −i

−i 7−→ i

←→ (1 2)(3 4), β :

√2 7−→ −√2

−√2 7−→ √2

i 7−→ i

−i 7−→ −i

←→ (1 2),

γ :

√2 7−→ √

2−√2 7−→ −√2

i 7−→ −i

−i 7−→ i

←→ (3 4).

In this description, the Galois group Gal(Q(ζ8)/Q) = Gal(Q(√

2, i)/Q) corresponds to thesubgroup

{id, (1 2), (3 4), (1 3)(2 4)} 6 S4.

While it can be hard to determine Galois groups in general, special arguments can sometimesbe exploited.

Example 4.9. Suppose that f(X) = X3 +aX2 + bX + c ∈ Q[X] is an irreducible cubic andthat f(X) has only one real root. Then Gal(Q(f(X))/Q) ∼= S3.

Proof. Let u1 ∈ R be the real root of f(X) and let u2, u3 be the remaining complexroots. Then Q(f(X)) = Q(u1, u2, u3) and in fact [Q(f(X)) : Q] = 6 since [Q(f(X)) : Q] | 6and u2 /∈ Q(u1) 6 R. Hence Gal(Q(f(X))/Q) is isomorphic to a subgroup of S3 and soGal(Q(f(X))/Q) ∼= S3 since the orders agree. We also have Q(f(X)) ∩ R = Q(u1).

The Galois group Gal(Q(f(X))/Q) contains an element of order 3 which corresponds to a3-cycle when viewed as a permutation of the roots u1, u2, u3; we can assume that this is (1 2 3).It also contains an element of order 2 obtained by restricting complex conjugation to Q(f(X));this fixes u1 and interchanges u2, u3, so it corresponds to the transposition (2 3). ¤

Remark 4.10. Such examples occur when the cubic polynomial f(X) has local maximumand minimum at real values c+ and c− with f(c+), f(c−) > 0 or f(c+), f(c−) < 0. This happensfor example with f(X) = X3− 3X +3 which has local extrema at ±1 and f(1) = 1, f(−1) = 5.


Given a Galois extension E/K, we will next study subextensions L/K 6 E/K and sub-groups Γ 6 Gal(E/K), focusing on the relationship between objects of these types.

4.3. Subgroups of Galois groups and their fixed fields

Let E/K a Galois extension and suppose that Γ 6 Gal(E/K). Consider the subset ofelements of E fixed by Γ,

EΓ = {u ∈ E : ∀γ ∈ Γ, γ(u) = u}.

Lemma 4.11. EΓ 6 E is a subfield of E containing K.

Proof. For u, v ∈ EΓ and γ ∈ Γ,

γ(u + v) = γ(u) + γ(v) = u + v, γ(uv) = γ(u)γ(v) = uv.

Also, if u 6= 0,

γ(u−1) = γ(u)−1 = u−1.

Finally, if t ∈ K then γ(t) = t, so K 6 EΓ. ¤

Definition 4.12. EΓ 6 E is the fixed subfield of Γ.

By Proposition 3.73, the extensions E/EΓ and EΓ/K are separable. E/EΓ is also normal,so this is itself a Galois extension. We will identify its Galois group. Notice that

[E : EΓ] = (E : EΓ) = |Gal(E/EΓ)|.

Now each element of Gal(E/EΓ) is also an element of Gal(E/K) and Gal(E/EΓ) 6 Gal(E/K).Notice that by definition Γ 6 Gal(E/EΓ), so Lagrange’s Theorem implies that |Γ| divides|Gal(E/EΓ)|. In fact we have

Proposition 4.13. For Γ 6 Gal(E/K), we have Gal(E/EΓ) = Γ and the equations

[E : EΓ] = |Gal(E/EΓ)| = |Γ|, [EΓ : K] =|Gal(E/K)|

|Γ| .

Proof. We know that E/EΓ is separable, so by the Primitive Element Theorem 3.75 it issimple, say E = EΓ(u). Now let the distinct elements of Γ be γ1 = id, γ2, . . . , γh, where h = |Γ|.Consider the polynomial of degree h

f(X) = (X − u)(X − γ2(u)) · · · (X − γh(u)) ∈ E[X].

Notice that f(X) is unchanged by applying any γk to its coefficients since the roots γj(u) arepermuted by γk. Hence, f(X) ∈ EΓ[X]. This shows that

[E : EΓ] = [EΓ(u) : EΓ] 6 h = |Γ|.

Since Γ 6 Gal(E/EΓ), we also have

h = |Γ| 6 |Gal(E/EΓ)| = [E : EΓ].

Combining these two inequalities we obtain

[E : EΓ] = |Gal(E/EΓ)| = |Γ| = h

and therefore Γ = Gal(E/EΓ). ¤

4.4. SUBFIELDS OF GALOIS EXTENSIONS AND RELATIVE GALOIS GROUPS 51

4.4. Subfields of Galois extensions and relative Galois groups

Let E/K a Galois extension and suppose that K 6 L 6 E. Then E/L is also a Galoisextension whose Galois group Gal(E/L) is sometimes called the relative Galois group of thepair of extensions E/K and L/K. The following is immediate.

Lemma 4.14. The relative Galois group of the pair of extensions E/K and L/K is a subgroupof Gal(E/K), i.e., Gal(E/L) 6 Gal(E/K), and its order is |Gal(E/L)| = [E : L].

Proposition 4.15. Let K 6 L 6 E. Then L = EGal(E/L).

Proof. Clearly L 6 EGal(E/L). Suppose that u ∈ E − L. Then there is an automorphismθ ∈ Gal(E/L) such that θ(u) 6= u, hence u /∈ EGal(E/L). This shows that EGal(E/L) 6 L andtherefore EGal(E/L) = L. ¤

We want to understand when Gal(E/L) 6 Gal(E/K) is actually a normal subgroup. Thenext result explains the connection between the two uses of the word normal which both derivefrom the Galois theoretic usage.

Proposition 4.16. Let E/K be a finite Galois extension and L/K 6 E/K a subextension.

i) The relative Galois group Gal(E/L) of the pair of extensions E/K and L/K is normalin Gal(E/K) if and only if L/K is normal.

ii) If L/K is normal and hence a Galois extension, then there is a group isomorphism

Gal(E/K)/Gal(E/L)∼=−→ Gal(L/K); α Gal(E/L) 7−→ α|L .

Proof. (i) Suppose that Gal(E/L)/Gal(E/K). Then if α ∈ Gal(E/L) and β ∈ Gal(E/K),we have βαβ−1 ∈ Gal(E/L). Now if u ∈ L, then for any γ ∈ Gal(E/K) and α ∈ Gal(E/L),γ(u) ∈ E satisfies

αγ(u) = γ(γ−1αγ(u)) = γ(u),

since γ−1αγ ∈ Gal(E/L). So L/K is normal.Conversely, if L/K is normal, then for every ϕ ∈ Gal(E/K) and v ∈ L, ϕ(v) ∈ L, so for

every θ ∈ Gal(E/L), θ(ϕ(v)) = ϕ(v) and therefore

ϕ−1θϕ(v) = v.

This shows that ϕ−1θϕ ∈ Gal(E/L). Hence for every ϕ ∈ Gal(E/K),

ϕGal(E/L)ϕ−1 = Gal(E/L),

which shows that Gal(E/L) / Gal(E/K).(ii) If α ∈ Gal(E/K), then αL = L since L/K is normal. Hence we can restrict α to an

automorphism of L,α|L : L −→ L; α|L(u) = α(u).

Then α|L is the identity function on L if and only if α ∈ Gal(E/L). It is easy to see that thefunction

Gal(E/K) −→ Gal(L/K); α 7−→ α|Lis a group homomorphism whose kernel is Gal(E/L). Thus we obtain an injective homomor-phism

Gal(E/K)/Gal(E/L) −→ Gal(L/K)

for which

|Gal(E/K)/Gal(E/L)| = [E : K][E : L]

= [L : K] = |Gal(L/K)|.Hence this must be an isomorphism of groups. ¤


4.5. The Galois Correspondence

We are now almost ready to state our central result which describes the Galois Correspon-dence associated with a finite Galois extension. We will use the following notation. For a finiteGalois extension E/K, let

S(E/K) = the set of all subgroups of Gal(E/K);

F(E/K) = the set of all subextensions L/K of L/K.

These sets are ordered by inclusion. Notice that since every subgroup of a finite group isequivalent to its underlying set, S(E/K) is a finite set. Define two functions by

ΦE/K : F(E/K) −→ S(E/K); ΦE/K(L) = Gal(E/L),

ΘE/K : S(E/K) −→ F(E/K); ΘE/K(Γ) = EΓ.

Theorem 4.17 (Main Theorem of Galois Theory). Let E/K be a finite Galois extension.Then the functions ΦE/K and ΘE/K are mutually inverse bijections which are order-reversing.

F(E/K)ΦE/K //

S(E/K)ΘE/K

oo

Under this correspondence, normal subextensions of E/K correspond to normal subgroups ofGal(E/K) and vice versa.

Proof. We know from Proposition 4.15 that for an extension L/K in F(E/K),

ΘE/K(ΦE/K(L)) = ΘE/K(Gal(E/L)) = EGal(E/L) = L.

Also, by Proposition 4.13 for H ∈ S(E/K) we have

ΦE/K(ΘE/K(Γ)) = ΦE/K(EΓ) = Gal(E/EΓ) = Γ.

This shows that ΦE/K and ΘE/K are mutually inverse and so are inverse bijections.Let L1/K, L2/K ∈ F(E/K) satisfy L1/K 6 L2/K. Then Gal(E/L2) 6 Gal(E/L1) since

L1 ⊆ L2 and so if α ∈ Gal(E/L2) then α fixes every element of L1. Hence ΦE/K(L2) 6 ΦE/K(L1)and so ΦE/K reverses order.

Similarly, if Γ1,Γ2 ∈ S(E/K) with Γ1 6 Γ2, then EΓ2 6 EΓ1 since if w ∈ EΓ2 then it isfixed by every element of Γ1 which is a subset of Γ2. Hence ΘE/K reverses order. ¤

There is an immediate consequence of the Main Theorem 4.17 which is closely related toProposition 3.20.

Corollary 4.18. Let E/K be a finite Galois extension. Then there are only finitely manysubextensions L/K 6 E/K.

Proof. Since the set S(E/K) is finite, so is F(E/K). ¤

When dealing with a finite Galois extension E/K, we indicate the subextensions in a diagramwith a line going upwards indicating an inclusion. We can also do this with the subgroups ofthe Galois group Gal(E/K) with labels indicating the index of the subgroups. In effect, theGalois Correspondence inverts these diagrams.

Example 4.19. Figure 4.1 displays the Galois Correspondence for the extension of Exam-ple 3.30.

As noted at the end of Example 3.38, the Galois group here is Gal(Q( 3√

2, ζ3)/Q) ∼= S3. Itis useful to make this isomorphism explicit. First take the 3 roots of the polynomial X3 − 2 forwhich E is the splitting field over Q; these are 3

√2, 3√

2 ζ3,3√

2 ζ23 which we number in the order

4.6. KAPLANSKY’S THEOREM 53

E = Q( 3√

2, ζ3)HH

¹¹

Q( 3√

2)NN

³³

2

eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

3

SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS Q( 3√

2 ζ3)NN

³³

2

kkkkkkkkkkkkkk

3

HHHHHHHHHHHHHHHHHHHHHHHHQ( 3√

2 ζ23 )

II

¸¸

2

3 Q(ζ3)QQ

°°

3

EEEEEEEEEEEEEEEEEEEEEE

2

kkkkkkkkkkkkkkkkkk

QSS

ΦE/K

®®Gal(E/Q)

Gal(E/Q(ζ3))

3

yyyyyyyyyyyyyyyyyyyyyy

2RRRRRRRRRRRRRR

Gal(E/Q( 3√

2))

2YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

3

kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

Gal(E/Q( 3√

2 ζ3))

2SSSSSSSSSSSSSSSSS

3

wwwwwwwwwwwwwwwwwwwwwww

Gal(E/Q( 3√

2 ζ23 ))

2

3

{id}

Figure 4.1. The Galois Correspondence for E = Q( 3√

2, ζ3)/Q

they are listed. Then the monomorphisms id, α0, α1, α′1, α2, α

′2 extend to automorphisms of E,

each of which permutes these 3 roots in the following ways given by cycle notation:

α0 = (2 3), α1 = (1 2 3), α′1 = (1 2), α2 = (1 3 2), α′2 = (1 3).

We find that

Gal(E/Q(ζ3)) = {id, α1, α2} ∼={id, (1 2 3), (1 3 2)}, Gal(E/Q( 3√

2)) = {id, α0} ∼={id, (2 3)},Gal(E/Q( 3

√2 ζ3)) = {id, α′2} ∼={id, (1 3)}, Gal(E/Q( 3

√2 ζ2

3 )) = {id, α′1} ∼={id, (1 2)}.Notice that {id, (1 2 3), (1 3 2)} / S3 and so Q(ζ3)/Q is a normal extension, hence it is a

splitting field. Of course it is the splitting field of X3 − 1 over Q.

4.6. Kaplansky’s Theorem

Theorem 4.20 (Kaplansky’s Theorem). Let f(X) = X4 + aX2 + b ∈ Q[X] be irreducible.

i) If b is a square in Q then Gal(Q(f(X))/Q) ∼= Z/2× Z/2.ii) If b(a2 − 4b) is a square in Q then Gal(Q(f(X))/Q) ∼= Z/4.iii) If neither b nor b(a2 − 4b) is a square in Q then Gal(Q(f(X))/Q) ∼= D8.

Proof. Let g(X) = X2 + aX + b ∈ Q[X]. Notice that g(X) must be irreducible sinceotherwise f(X) would factorize, hence (a2− 4b) is not a square in Q. Setting d = (a2− 4b) ∈ Qand δ to be a square root of d, we see that the roots of g(X) are (−a± δ)/2, where δ /∈ Q. Then


the roots of f(X) are ±u,±v, where

u2 =(−a + δ)

2, v2 =

(−a− δ)2

,

so the splitting field of f(X) over Q is E = Q(u, v) which contains the quadratic extensionQ(δ)/Q. Since deg f(X) = 4, we must also have 4 | [E : Q]. In fact, since E is obtained by atmost 3 successive quadratic extensions we also have [E : Q] | 8.

(i) We have

(uv)2 = u2v2 =a2 − d

4=

4b

4= b,

hence uv is a square root of b which is in Q. Setting c = uv ∈ Q, we find that v = c/u ∈ Q(u).This shows that E = Q(u) and we have the following Galois tower.

E = Q(u)

2

Q(δ)

2

Q

In particular [E : Q] = 4 = |Gal(E/Q)|. Notice that for the Galois extension Q(δ)/Q theremust be a normal subgroup N / Gal(E/Q) with

Q(δ) = EN , Gal(Q(δ)/Q) = Gal(E/Q)/N.

Hence there is an element σ ∈ Gal(E/Q) for which σ(δ) = −δ. This element must also havethe effects σ(u) = ±v and σ(v) = ±u. Given u we might as well choose v so that σ(u) = v.There is also an element τ ∈ N for which τ(u) = −u and we also have τ(v) = −v. Notice thatif σ(v) = −u then easy calculation shows that

τσ(v) = στ(v) = u, τσ(δ) = στ(δ) = −δ,

hence we might as assume that σ(v) = u since if necessary we can replace our original choiceby τσ.

We now haveσ(u) =

c

u, τ(u) = −u, τσ(u) = στ(u) = − c

u.

These satisfyσ2 = τ2 = (στ)2 = id = the identity, στ = τσ.

This shows that

Gal(Q(f(X))/Q) = Gal(E/Q) = {id, σ, τ, στ} ∼= Z/2× Z/2 = a Klein 4-group.

(ii) If bd is a square in Q, then

(uvδ)2 = u2v2d = bd,

which is a square in Q, so we can write uvδ = c ∈ Q or equivalently v = c/(uδ) ∈ Q(u) sinceQ(δ) 6 Q(u). This shows that E = Q(u, v) = Q(u) and again we have a Galois tower

E = Q(u)

2

Q(δ)

2

Q

4.6. KAPLANSKY’S THEOREM 55

with [E : Q] = 4 = |Gal(E/Q)|.Since Q(δ)/Q is Galois there is an element σ ∈ Gal(E/Q with σ(δ) = −δ and this has the

effect σ(u) = ±v; given u we might as well choose v so that σ(u) = v. Notice that

σ(v) =c

σ(uδ)= − c

vδ= −u,

so σ2(u) = −u. This shows that

Gal(Q(f(X))/Q) = Gal(E/Q) = {id, σ, σ2σ3} ∼= Z/4 = a cyclic group of order 4.

(iii) Suppose that d, b and bd are not squares in Q. By an easy calculation we find that(uv)2 = b, so uv ∈ E is a square root of b in E. Suppose that uv ∈ Q(δ); then uv = p + qδ forsome p, q ∈ Q. By squaring we obtain

b = (p2 + q2d) + 2pqδ,

and so pq = 0. We cannot have q = 0 since this would imply that b was a square in Q; ifp = 0 then b = q2d and so bd = (qd)2, implying that bd was a square in Q. Thus we haveQ(uv) ∩Q(δ) = Q. A similar discussion shows that

Q(uvδ) ∩Q(δ) = Q = Q(uvδ) ∩Q(uv).

So we have a Galois tower which includes the following subfields.

E = Q(u, v)

Q(uv, δ)

rrrrrrrrrr

MMMMMMMMMMM

Q(δ)

2 LLLLLLLLLLLLQ(uv)

2

Q(uvδ)

2pppppppppppp

Q

Chooseα ∈ Gal(E/Q(uv)) 6 Gal(E/Q)

so that α(δ) = −δ. By renaming −v to v if necessary, we may assume that v = α(u) and sou = α(v). Notice that α2 = id.

Chooseβ ∈ Gal(E/Q(δ)) 6 Gal(E/Q)

with β(uv) = −uv. We must have either β(u) = −u or β(v) = −v, so by interchanging ±δ ifnecessary we can assume that β(u) = −u and β(v) = v. Notice that β2 = id.

Chooseγ ∈ Gal(E/Q(δ, uv)) 6 Gal(E/Q)

so that γ(u) = −u. Then we must have γ(v) = −v since γ(uv) = uv. Notice that γ2 = id.Setting σ = αβ we find σ(u) = −v and σ(v) = u. Then σ2 = γ and σ has order 4. Also,

ασα = βσβ = σ−1.

The eight elementsid, σ, γ, σ−1, α, ασ, αγ, ασ−1

form a group isomorphic to the dihedral group of order 8, D8. Therefore we have

Gal(Q(f(X))/Q) = Gal(E/Q) ∼= D8,


and [E : Q] = 8. The corresponding Galois tower is

E = Q(u, v)

2

Q(uv, δ)2

rrrrrrrrrr2

2

MMMMMMMMMMM

Q(δ)

2 LLLLLLLLLLLLQ(uv)

2

Q(uvδ)

2pppppppppppp

Q

¤

Example 4.21. We have the following Galois groups:

Gal(Q(X4 + 1)/Q) ∼= Z/2× Z/2; Gal(Q(X4 + 4X2 + 2)/Q) ∼= Z/4;

Gal(Q(X4 + 2X2 + 2)/Q) ∼= D8 .

4.7. Galois extensions inside the complex numbers

Let E/Q be a finite Galois extension with E/Q 6 C/Q. Setting ER = R ∩ E, we haveQ 6 ER 6 E.

Proposition 4.22. Complex conjugation ( ) : C −→ C restricts to an automorphism of E

over Q, ( )E/Q : E −→ E.

i) ( )E/Q which agrees with the identity function if and only if ER = E.ii) If ER 6= E then ⟨

( )E/Q⟩

= {id, ( )E/Q} ∼= Z/2,

hence ER = E〈( )E/Q〉 and [E : ER] = 2.

Proof. Let u ∈ E. As E/Q is normal, minpolyQ,u(X) ∈ Q[X] splits over E, hence all ofits complex roots lie in E. But ( ) permutes the roots of this polynomial. Hence ( ) maps E

into itself.(i) For z ∈ C, z = z if and only if z ∈ R.(ii) Here | ⟨( )E/Q

⟩ | = 2, and

E〈( )E/Q〉 = {u ∈ E : u = u} = ER. C2

∞R

∞ E

2nnnnnnnnnnnnnnn

²²²²²²²²²²²²²²²

ER

AAAA

AAAA

Q

¤

We will usually write ( ) rather than ( )E/Q when no confusion seems likely to result.

4.7. GALOIS EXTENSIONS INSIDE THE COMPLEX NUMBERS 57

Example 4.23. Consider the cyclotomic extension Q(ζ8)/Q where

ζ8 = eπi/4 =1√2

+1√2i.

From Example 4.8 we know that

Q(ζ8) = Q(√

2, i), [Q(ζ8) : Q] = 4,

and we easily see thatQ(ζ8)R = Q(

√2).



4.1. If f(X) ∈ K[X] is a separable polynomial, prove that the splitting field of f(X) over K isa finite Galois extension of K.

4.2. Let K be a field for which charK 6= 2, 3 and suppose that f(X) ∈ K[x] is a cubicpolynomial.

(a) Show that there u, v ∈ K with u 6= 0 such that f(uX + v) = X3 + aX + b for somea, b ∈ K. If f(X) is monic, deduce that a, b ∈ K; under what conditions is this alwaystrue?

(b) If g(X) = X3 + aX + b ∈ K[x] is irreducible and E = K(g(X)) is its splitting fieldover K, explain why Gal(E/K) is isomorphic to one of the groups S3 or A3.

(c) Continuing with the notation and assumptions of (b), suppose that w1, w2, w3 are thedistinct roots of g(X) in E and let

∆ = (w1 − w2)2(w3 − w3)2(w1 − w3)2 ∈ E.

Show that∆ = −4b3 − 27a2,

and hence ∆ ∈ K. If δ = (w1 − w2)(w3 − w3)(w1 − w3), show that

Gal(E/K) ∼={

A3 if δ ∈ K,

S3 if δ /∈ K.

[Hint: Consider K(δ) 6 E and the effect of even and odd permutations in Gal(E/K) 6S3 on the element δ.]

4.3. This is a revision exercise on finite groups of small order.(a) Show that every non-abelian finite group has order at least 6.(b) Let D8 be the dihedral group with the eight elements

ι, α, α2, α3, β, βα, βα2, βα3

satisfyingα4 = ι, β2 = ι, βαβ = α−1 = α3.

Find all the normal subgroups of D8.

4.4. Use Kaplansky’s Theorem 4.20 to find the Galois group of the splitting field E of thepolynomial X4 + 3 ∈ Q[X] over Q. Determine all the subextensions F 6 E for which F/Q isGalois.

4.5. Find the Galois groups for each of the following extensions:Q(X3 − 10)/Q; Q(

√2)(X3 − 10)/Q(

√2); Q(

√3 i)(X3 − 10)/Q(

√3 i);

Q(√

23 i)(X3 −X − 1)/Q(√

23 i); K(X3 −X − 1)/K for K = Q, Q(√

5), Q(√

5 i), Q(i).

CHAPTER 5

Galois extensions in positive characteristic

In this chapter we will investigate extensions of fields of positive characteristic, especiallyfinite fields. A thorough account of finite fields can be found in [4]. We will assume that p > 0is a prime and K is a field of characteristic charK = p; we will also assume that K containsthe prime subfield Fp, i.e., Fp 6 K.

5.1. Finite fields

If K is a finite field, then K is an Fp-vector space. Our first goal is to count the elements ofK. Here is a more general result.

Lemma 5.1. Let F be a finite field with q elements and let V be an F -vector space. ThendimF V < ∞ if and only if V is finite in which case |V | = qdimF V .

Proof. If d = dimF V < ∞, then for a basis v1, . . . , vd we can express each element v ∈ V

uniquely in the form v = t1v1 + · · · + tdvd, where t1, . . . , td ∈ F . Clearly there are exactly qd

such expressions, so |V | = qd.Conversely, if V is finite then any basis has finitely many elements and so dimF V < ∞. ¤

Corollary 5.2. Let F be a finite field and E/F an extension. Then E is finite if and onlyif E/F is finite and then |E| = |F |[E:F ].

Corollary 5.3. Let K be a finite field. Then K/Fp is finite and |K| = p[K:Fp].

Our next task is to show that for each power pd there is a finite field with pd elements. Westart with the algebraic closure Fp of Fp and consider the polynomial Θpd(X) = Xpd − X ∈Fp[X]. Notice that Θ′

pd(X) = −1, hence by Proposition 3.55 every root of Θpd(X) in Fp issimple. Therefore by Corollary 1.31 Θpd(X) must have exactly pd distinct roots in Fp, say0, u1, . . . , upd−1. Then in Fp[X] we have

Xpd −X = X(X − u1) · · · (X − upd−1),

and each root is separable over Fp. Let

Fpd = {u ∈ Fp : Θpd(u) = 0} ⊆ Fp, F0pd = {u ∈ Fpd : u 6= 0}.

Notice that u ∈ F0pd if and only if upd−1 = 1.

Proposition 5.4. For each d > 1, Fpd is a finite subfield of Fp with pd elements andF0

pd = F×pd. Furthermore, the extension Fpd/Fp is separable.

Proof. If u, v ∈ Fpd then by the Idiot’s Binomial Theorem 1.8,

(u + v)pd − (u + v) = (upd+ vpd

)− (u + v) = (upd − u) + (vpd)− v) = 0,

(uv)pd − uv = updvpd − uv = uv − uv = 0.

Furthermore, if u 6= then upd−1 = 1 and so u has multiplicative inverse upd−2. Hence Fpd 6 Fp.Notice that Fp 6 Fpd , so Fpd/Fp is a finite extensions. In any field the non-zero elements arealways invertible, hence F0

pd = F×pd . ¤

59

60 5. GALOIS EXTENSIONS IN POSITIVE CHARACTERISTIC

Definition 5.5. The finite subfield Fpd 6 Fp is called the Galois field of order pd and issometimes also denoted GF(pd).

Of course, Fp1 = GF(p1) = Fp. Notice also that [Fpd : Fp] = d.

Proposition 5.6. Let d > 1.

i) Fpd 6 Fp is the splitting subfield for each of the polynomials Xpd −X and Xpd−1 − 1over Fp.

ii) Fpd 6 Fp is the unique subfield with pd elements.iii) If K is any field with pd elements then there is an monomorphism K −→ Fp with image

Fpd, hence K ∼= Fpd.

Proof. (i) As Fpd consists of exactly the roots of Θpd(X) in Fp, it is the splitting subfield.The non-zero elements of Fpd are the roots of Xpd−1− 1, so Fpd is also the splitting subfield forthis polynomial.(ii) Let K 6 Fp have pd elements. Notice that the non-zero elements of F form a group K×

under multiplication. This group is abelian and has pd−1 elements, so by Lagrange’s Theorem,each element u ∈ K× has order dividing pd − 1, therefore upd−1 = 1 and so upd

= u. But thismeans every element of K is a root of Θpd(X) and so K 6 Fpd ; equality follows since thesesubfields both have pd elements.(iii) Apply the Monomorphism Extension Theorem 3.49 for K = E = Fp and L = K. ¤

It is worth noting the following consequence of this result and the construction of Fpd .

Corollary 5.7. Let K be a finite field of characteristic p. Then the extension K/Fp isseparable.

Example 5.8. Consider the polynomial X4 −X ∈ F2[X]. By inspection, in the ring F2[X]we find that

X4 −X = X4 + X = X(X3 + 1) = X(X + 1)(X2 + X + 1).

Now X2 + X + 1 has no root in F2 so it must be irreducible in F2[X]. It splitting field is aquadratic extension F2(w)/F2 where w is one of the roots of X2 + X + 1, the other being w + 1since the sum of the roots is the coefficient of X. This tells us that every element of F4 = F2(w)can be uniquely expressed in the form a + bw with a, b ∈ F2. To calculate products we use thefact that w2 = w + 1, so for a, b, c, d ∈ F2 we have

(a + bw)(c + dw) = ac + (ad + bc)w + bdw2 = (ac + bd) + (ad + bc + bd)w.

Example 5.9. Consider the polynomial X9 −X ∈ F3[X]. Let us find an irreducible poly-nomial of degree 2 in F3[X]. Notice that X2 + 1 has no root in F3, hence X2 + 1 ∈ F3[X] isirreducible; so if u ∈ F3 is a root of X2 + 1 then F3(u)/F3 has degree 2 and F3(u) = F9. Everyelement of F9 can be uniquely expressed in the form a + bu with a, b ∈ F3. Multiplication iscarried out using the relation u2 = −1 = 2.

By inspection, in the ring F3[X] we find that

X9 −X = X(X8 − 1) = (X3 −X)(X2 + 1)(X2 + X − 1)(X2 −X − 1).

So X2 +X− 1 and X2−X− 1 are also quadratic irreducibles in F3[X]. We can find their rootsin F9 using the quadratic formula since in F3 we have 2−1 = (−1)−1 = −1. The discriminant ofX2 + X − 1 is

1− 4(−1) = 5 = 2 = u2,

so its roots are (−1)(−1± u) = 1± u. Similarly, the discriminant of X2 −X − 1 is

1− 4(−1) = 5 = 2 = u2

5.1. FINITE FIELDS 61

and its roots are (−1)(1± u) = −1± u. Then we have

F9 = F3(u) = F3(1± u) = F3(−1± u).

There are two issues we can now clarify.

Proposition 5.10. Let Fpm and Fpn be two Galois fields of characteristic p. Then Fpm 6Fpn if and only if m | n.

Proof. If Fpm 6 Fpn , then by Corollary 5.2,

pn = (pm)[Fpn :Fpm ],

so m | n.If m | n, write n = km with k > 1. Then for u ∈ Fpm we have upm

= u, so

upn= upmk

= (upm)pm(k−1)

= upm(k−1)= · · · = upm

= u.

Hence u ∈ Fpn and therefore Fpm 6 Fpn . ¤

This means that we can think of the Galois fields Fpn as ordered by divisibility. Here is thediagram of subfields for Fp24 showing extensions with no intermediate subextensions.

(5.1) Fp24

{{{{

{{{{

Fp8 Fp12

{{{{

{{{{

Fp4 Fp6

{{{{

{{{{

CCCC

CCCC

Fp2

CCCC

CCCC

Fp3

{{{{

{{{{

Fp

Theorem 5.11. The algebraic closure of Fp is the union of all the Galois field of character-istic p,

Fp =⋃

n>1

Fpn .

Furthermore, each element u ∈ Fp is separable over Fp.

Proof. Let u ∈ Fp. Then u is algebraic over Fp and the extension Fp(u)/Fp is finite. Henceby Corollary 5.2, Fp(u) 6 Fp is a finite subfield. Proposition 5.10 now implies that Fp(u) = Fpn

for some n. The separability statement follows from Corollary 5.7. ¤

We will require a useful fact about Galois fields.

Proposition 5.12. The group of units F×pd in Fpd is cyclic.

This is a special case of a more general result about arbitrary fields.

Proposition 5.13. Let K be a field. Then every finite subgroup U 6 K× is cyclic.

Proof. Use Corollary 1.31 and Lemma 1.41. ¤

Definition 5.14. w ∈ F×pd is called a primitive root if it is a primitive (pd − 1)-th root of

unity, i.e., its order in the group F×pd is (pd − 1), hence 〈w〉 = F×

pd .


Remark 5.15. Unfortunately the word primitive has two confusingly similar uses in thecontext of finite fields. Indeed, some authors use the term primitive element for what we havecalled a primitive root, but that conflicts with our usage, although as we will in the next result,every primitive root is indeed a primitive element in our sense!

Proposition 5.16. The extension of Galois fields Fpnd/Fpd is simple, i.e., Fpnd = Fpd(u)for some u ∈ Fpnd.

Proof. By Proposition 5.12, Fpnd has a primitive root w say. Then every element of Fpnd

is a polynomial in w, so Fpnd 6 Fpd(w) 6 Fpnd , hence Fpnd = Fpd(w). ¤

Remark 5.17. This completes the proof of the Primitive Element Theorem 3.75 which wehad previously only established for infinite fields.

Example 5.18. In Example 5.8 we find that F4 = F2(w) has the two primitive roots w andw + 1.

Example 5.19. In Example 5.9 we have F9 = F3(u) and F×9 is cyclic of order 8. Sinceϕ(8) = 4, there are four primitive roots and these are the roots of the polynomials X2 + X − 1and X2 −X − 1 which we found to be ±1± u.

We record a fact that is very important in Number Theory.

Proposition 5.20. Let p > 0 be an odd prime.i) If p ≡ 1 mod 4, the polynomial X2 + 1 ∈ Fp[X] has two roots in Fp.ii) If p ≡ 3 mod 4 the polynomial X2 +1 ∈ Fp[X] is irreducible, so Fp2

∼= Fp[X]/(X2 +1).

Proof. (i) We have 4 | (p− 1) = |F×p |, so if u ∈ F×p is a generator of this cyclic group, theorder of u|F

×p |/4 is 4, hence this is a root of X2 + 1 (the other root is −u|F

×p |/4).

(ii) If v ∈ Fp is a root of X2 + 1 then v has order 4 in F×p . But then 4 | (p− 1) = |F×p |, which isimpossible since p− 1 ≡ 2 mod 4. ¤

Here is a generalization of Proposition 5.20.

Proposition 5.21. Fpd contains a primitive n-th root of unity if and only if pd ≡ 1 mod n

and p - n.

5.2. Galois groups of finite fields

Consider an extension of Galois fields Fpnd/Fpd . By Proposition 5.6(i), Corollary 5.7 andProposition 3.73, this extension is Galois and

|Gal(Fpnd/Fpd)| = [Fpnd : Fpd ] = n.

We next introduce an important element of the Galois group Gal(Fpnd/Fpd).

Definition 5.22. The (relative) Frobenius mapping for the extension Fpnd/Fpd is the func-tion Fd : Fpnd −→ Fpnd given by Fd(t) = tp

d.

Proposition 5.23. The relative Frobenius mapping Fd : Fpnd −→ Fpnd is an automor-phism of Fpnd that fixes the elements of Fpd, so Fd ∈ Gal(Fpd/Fpnd). The order of Fd is n,so Gal(Fpd/Fpnd) = 〈Fd〉, the cyclic group generated by Fd.

Proof. For u, v ∈ Fpnd , we have the identities

Fd(u + v) = (u + v)pd= upd

+ vpd, Fd(uv) = (uv)pd

= updvpd

,

so Fd is a ring homomorphism. Also, for u ∈ Fpd we have

Fd(u) = upd= u,

5.2. GALOIS GROUPS OF FINITE FIELDS 63

so Fd fixes the elements of Fpd . To see that Fd is an automorphism, notice that the compositionpower Fn

d = Fd ◦ · · · ◦ Fd (with n factors) satisfies

Fnd(t) = tp

nd= t

for all t ∈ Fpnd , hence Fnd = id. Then Fd is invertible with inverse F−1

d = Fn−1d . This also shows

that the order of Fd in the group AutFpd

(Fpnd) is at most n. Suppose the order is k with k 6 n;

then every element u ∈ Fpnd satisfies the equation Fkd(u) = u which expands to upkd

= u, henceu ∈ Fpkd . But this can only be true if k = n. ¤

Actually, Frobenius mappings exists on the algebraic closure Fp. For d > 1, consider thefunction

Fd : Fp −→ Fp ; Fd(t) = tpd.

The proof of our next result is left as an exercise.

Proposition 5.24. Let d > 1.

i) The Frobenius mapping Fd : Fp −→ Fp is an automorphism of Fp which fixes the ele-ments of Fpd. In fact for u ∈ Fp, Fd(u) = u if and only if u ∈ Fpd.

ii) The restriction of Fd to a Galois subfield Fpdn agrees with the relative Frobenius map-ping Fd : Fpnd −→ Fpnd.

ii) If k > 1, then Fkd = Fkd, so Fd has infinite order in the automorphism group AutF

pd(Fp),

hence this group is infinite.

The Frobenius mapping F = F1 is often called the absolute Frobenius mapping since it existsas an element of each of the groups AutFp(Fp) and AutFp(Fpn) = Gal(Fpn/Fp) for every n > 1.

In Gal(Fpnd/Fpd) = 〈Fd〉, for each k with k | n there is the cyclic subgroup⟨Fk

d

⟩of order

| ⟨Fkd

⟩ | = n/k.

Proposition 5.25. For k | n, the fixed subfield of⟨Fk

d

⟩in Fpnd is F〈F

kd〉

pnd = Fpdk .

Fpnd

n/k

F〈Fkd〉

pnd = Fpdk

k

Fpd

Proof. For u ∈ Fpnd we have Fkd(u) = updk

, hence Fkd(u) = u if and only if u ∈ Fpdk . ¤

Here is the subgroup diagram corresponding to the lattice of subfields of Fp24 shown in (5.1).


(5.2) Gal(Fp24/Fp) = 〈F〉 ∼= Z/24

⟨F2

⟩

lllllllllllllll ⟨F3

⟩

RRRRRRRRRRRRRRR

⟨F4

⟩ ⟨F6

⟩

RRRRRRRRRRRRRRRRR

lllllllllllllllll

⟨F8

⟩ ⟨F12

⟩

RRRRRRRRRRRRRRRRR

⟨F24

⟩= {id}

RRRRRRRRRRRRRRR

5.3. The trace and norm mappings

For an extension of Galois fields Fpnd/Fpd , consider the function TrFpnd/F

pd: Fpnd −→ Fpnd

defined by

TrFpnd/F

pd(u) = u + upd

+ up2d+ · · ·+ up(n−1)d

= u + Fd(u) + F2d(u) + · · ·+ F(n−1)d(u).

Notice that

Fd(TrFpnd/F

pd(u)) = upd

+ up2d+ up3d

+ · · ·+ upnd

= upd+ up2d

+ up3d+ · · ·+ up(n−1)d

+ u

= TrFpnd/F

pd(u).

So by Proposition 5.24(i), TrFpnd/F

pd(u)) ∈ Fpd . Therefore we can redefine this mapping to have

codomain Fpd , giving the relative trace TrFpnd/F

pd: Fpnd −→ Fpd .

Proposition 5.26. The relative trace TrFpnd/F

pdis a surjective Fpd-linear mapping and

whose kernel is an Fpd-vector subspace of dimension n− 1.

Proof. Clearly TrFpnd/F

pdis additive. For t ∈ Fpd we have tp

d= t, so linearity follows from

the formula

tu + (tu)pd+ (tu)p2d

+ · · ·+ (tu)p(n−1)d= tu + tupd

+ tup2d+ · · ·+ tup(n−1)d

.

To see that TrFpnd/F

pdis surjective, notice that TrF

pnd/Fpd

(u) = 0 if and only if u is a root ofthe polynomial

X + Xpd+ Xp2d

+ · · ·+ Xp(n−1)d ∈ Fpd [X]

which has degree p(n−1)d and so has at most p(n−1)d < pnd roots in Fpnd . This means thatkerTrF

pnd/Fpd

cannot be the whole of Fpnd . TrFpnd/F

pdis surjective since its codomain has

dimension 1. ¤

A multiplicative version of this construction can also be defined. Consider the function

NFpnd/F

pd: F×

pnd −→ F×pd ; NF

pnd/Fpd

(u) = uupdup2d · · ·up(n−1)d

= uFd(u) F2d(u) · · ·F(n−1)d(u).

5.3. THE TRACE AND NORM MAPPINGS 65

Then we have

Fd(NFpnd/F

pd(u)) = upd

up2dup3d · · ·upnd

= updup2d

up3d · · ·up(n−1)du

= uupdup2d

up3d · · ·up(n−1)d

= NFpnd/F

pd(u).

So by Proposition 5.24(i), NFpnd/F

pd(u) ∈ Fpd . By redefining the codomain we obtain the relative

normNormF

pnd/Fpd

: F×pnd −→ F×

pd ; NormFpnd/F

pd(u) = uupd

up2d · · ·up(n−1)d.

Proposition 5.27. The relative norm NormFpnd/F

pdis a surjective group homomorphism.

Proof. Multiplicativity is obvious. The kernel of NormFpnd/F

pdconsists of the roots in Fpnd

of the polynomialX1+pd+···+p(n−1)d − 1 ∈ Fpd [X],

so

| kerNormFpnd/F

pd| 6 1 + pd + · · ·+ p(n−1)d =

pnd − 1pd − 1

.

Hence

| imNormFpnd/F

pd| = pnd − 1

| kerNormFpnd/F

pd| > pd − 1.

Since imNormFpnd/F

pd6 F×

pd , we also have

| imNormFpnd/F

pd| 6 pd − 1,

thereforeim NormF

pnd/Fpd

= F×pd . ¤



5.1. Show that Proposition 5.13 also applies to an integral domain in place of a field.

5.2. What happens to Theorem 5.20 if we try to take p = 2.

5.3. Let f(X) ∈ Fpd [X] be an irreducible polynomial with deg f(X) = n. Find the splitting fieldof f(X). Deduce that for any other irreducible polynomial g(X) ∈ Fpd [X] with deg g(X) = n,the splitting fields of f(X) and g(X) over Fpd agree.

5.4. Find the smallest Galois fields containing all the roots of the following polynomials:(a) X8 − 1 ∈ F41[X]; (b) X8 − 1 ∈ F5[X]; (c) X8 − 1 ∈ F11[X]; (d) X8 − 1 ∈ F2[X].In each case find a primitive root of this Galois field.

5.5. Let w ∈ F×pd be a primitive root. If ` < d, show that w /∈ F×

p` . Deduce that degFpw = d

and d | ϕ(pd − 1).

5.6. Let p > 0 be a prime. Suppose that d > 1, and K/Fpd is an extension. For a ∈ K, letga(X) = Xpd −X − a ∈ K[X].

(a) If the polynomial ga(X) is irreducible over K, show that the splitting field E of ga(X)over K is separable and Gal(E/K) ∼= Fpd . [Hint: show that if u ∈ E is a root of ga(X)in an extension E/K, then so is u + t for every t ∈ Fp.]

(b) If d = 1, show that ga(X) is irreducible over K if and only if it has no root in K.(c) If K is a finite field and d > 1, explain why ga(X) can never be irreducible over K.

CHAPTER 6

A Galois Miscellany

In this chapter we will explore some miscellaneous topics in Galois Theory. Historically,Galois Theory has always been an important tool in Number Theory and Algebra, stimulatingthe development of subjects such as Group Theory, Ring Theory and such diverse areas asDifferential Equations, Complex Analysis and Algebraic Geometry. Many of the ideas we willmeet in this chapter are of importance in such mathematical areas.

6.1. A proof of the Fundamental Theorem of Algebra

We will prove the Fundamental Theorem of Algebra for the complex numbers C. This proofis essentially due to Gauss but he did not use (or know) Sylow theory. It is interesting tocompare our proof with others which use the topology of the plane and circle; our proof onlyuses the connectivity of the real line together with explicit algebraic calculations in C.

Theorem 6.1 (The Fundamental Theorem of Algebra). The field of complex numbers C isalgebraically closed and R = C.

Proof. We know that [C : R] = 2, so C/R is algebraic. Let p(X) ∈ C[X] be irre-ducible. Then any root u of p(X) in the algebraic closure C is algebraic over R, so p(X) |minpolyR,u(X) in C[X]. The splitting field of p(X) over C is contained in the splitting field E

of minpolyR,u(X)(X2 + 1) over R. Since C 6 E, we have 2 | [E : R] and so Gal(E/R).Now consider a 2-Sylow subgroup P 6 Gal(E/R). The fixed field of P has odd degree over

R since

[EP : R] =|Gal(E/R)|

|P | = odd.

So we can consider the odd degree extension EP /R. Now the Primitive Element Theorem 3.75allows us to write EP = R(v) for some v whose minimal polynomial (over R) has odd degree.But by the Intermediate Value Theorem, every real polynomial of odd degree has a real root,so irreducibility implies that v has degree 1 over R and therefore EP = R. This shows thatGal(E/R) = P , therefore Gal(E/R) is a 2-group.

As C/R is a Galois extension, we can consider the normal subgroup Gal(E/C)/Gal(E/R) forwhich |Gal(E/R)| = 2|Gal(E/C)|. We must show that |Gal(E/C)| = 1, so suppose not. Fromthe theory of 2-groups, there is a normal subgroup N /Gal(E/C) of index 2, so we can considerthe Galois extension EN/C of degree 2. But from known properties of C (see Proposition 3.29),every quadratic aX2 + bX + c ∈ C[X] has complex roots (basically because we can find squareroots of every complex number). So we cannot have an irreducible quadratic polynomial inC[X]. Therefore |Gal(E/C)| = 1 and so E = C. ¤

6.2. Cyclotomic extensions

We begin by recording the situation of cyclotomic extensions over Q. We will use materialdiscussed in Section 1.3. Let ζn = e2πi/n, the standard primitive n-th root of 1 in C. InTheorem 1.38, it was claimed that the irreducible polynomial over Q which has ζn as a root was

67

68 6. A GALOIS MISCELLANY

the n-th cyclotomic polynomial

Φn(X) =∏

t=1,...,n−1gcd(t,n)=1

(X − ζtn).

Theorem 6.2. Let n > 2. Then

• Q(ζn) = Q[X]/(Φn(X));• [Q(ζn) : Q] = ϕ(n);• Gal(Q(ζn)/Q) ∼= (Z/n)×, where the element tn ∈ (Z/n)× acts on Q(ζn) by tn · ζn = ζt

n.

Proof. Since the complex roots of Φn(X) are the powers ζtn with t = 1, . . . , n − 1 and

gcd(t, n) = 1, Q(ζn) is the splitting field of Φn(X) over Q and indeed Q(ζn) = Q(ζtn) whenever

t has the above properties and so ζtn is a primitive n-th root of unity. The main step in the

proof is to show that Φn(X) ∈ Z[X] is irreducible. To do this we will show that every power ζtn

as above is actually a Galois conjugate of ζn over Q, hence

Φn(X) = minpolyQ,ζn(X) = minpolyQ,ζt

n(X)

and so is irreducible.Consider

Z(ζn) = {a0 + a1ζn + · · ·+ arζrn : r > 0, aj ∈ Z} ⊆ Q(ζn).

Then Z(ζn) is a subring of Q(ζn) and so is an integral domain. Its group of units contains thecyclic subgroup 〈ζn〉 of order n.

Let p > 0 be a prime which does not divide n. Let P / Z(ζn) be a maximal ideal whichcontains p; then the quotient ring Z(ζn)/P is a field of characteristic p. In fact, it is a finitefield, say Fpd for some d. Let π : Z(ζn) −→ Fpd be the quotient homomorphism.

Inside the group of units of Z(ζn) is the subgroup of powers of ζn, 〈ζn〉 6 Z(ζn)×; thisis a cyclic subgroup of order n. We claim that when restricted to 〈ζn〉, π gives an injectivegroup homomorphism, π′ : 〈ζn〉 −→ F×

pd . To see this, suppose that π′(ζrn) = 1 for some r =

1, 2, . . . , n− 1; then ζrn − 1 ∈ P . By elementary Group Theory we can assume that r | n and so

p - r. Factoring, we have

(ζn − 1)(ζr−1n + · · ·+ ζn + 1) ≡ (ζn − 1)r mod P,

so ζn − 1 ∈ P or r ∈ P since maximal ideals are prime. But Z ∩ P = (p) and so r /∈ P , henceζn − 1 ∈ P . Recalling that

ζn−1n + · · ·+ ζn + 1 = 0,

we see that n ∈ P and hence p | n, thus contradicting our original assumption on n. So π′ isinjective.

Writing u = π′(u), we can consider the effect of the absolute Frobenius map F: Fpd −→ Fpd

on ζtn = ζt

n,

F(ζtn) = (ζt

n)p = ζtpn .

This shows that in the Galois extension Fpd/Fp, ζtn is conjugate to ζtp

n ; by iterating this we find

that ζtn is conjugate to every power of the form ζtpk

n .Now let t = 1, . . . , n− 1 and gcd(t, n) = 1. Suppose that there is a factorization

Φn(X) = f(X)minpolyQ,ζn(X)

for some monic polynomial f(X) ∈ Z[X] and f(ζtn) = 0. Consider the prime power factorization

t = pr11 · · · prm

m , where the pj are primes with 2 6 p1 < · · · < pm and rj > 1 with. Notice thatpj - n since gcd(t, n) = 1.

6.2. CYCLOTOMIC EXTENSIONS 69

Now consider a maximal ideal P1 / Z[ζn] containing p1. Reducing modulo P1 and working

in the resulting extension Fp

d11

/Fp1 , we find that ζn is conjugate to ζp

r11

n . By separability and

the fact that the reduction map π1 : Z[ζn] −→ Fp

d11

is injective on the powers of ζn, we find that

f(ζpr11

n ) 6= 0 and so f(ζpr11

n ) 6= 0 in Z[ζn]. This shows that

minpolyQ,ζn(ζp

r11

n ) = 0

and so ζp

r11

n is conjugate to ζn.Repeating this argument starting with ζ

pr11

n and using the prime p2 we find that

minpolyQ,ζn(ζp

r11 p

r22

n ) = 0

and so ζp

r11 p

r22

n is conjugate to ζn. Continuing in this fashion, for each j = 1, . . . , m we obtain

minpolyQ,ζn(ζ

pr11 p

r22 ···prj

jn ) = 0,

and so ζp

r11 ···prj

jn is conjugate to ζn. When j = m, this shows that minpolyQ,ζn

(ζtn) = 0. Hence

ζtn is conjugate to ζn in the extension Q(ζn)/Q. ¤

Proposition 6.3. For n > 2, consider the cyclotomic extension Q(ζn)/Q where ζn = e2πi/n.Then Q(ζn)R 6= Q(ζn). Furthermore,

Q(ζn)R = Q(ζn)〈( )〉 = Q(ζn + ζn) = Q(cos(2π/n)),

and

[Q(cos(2π/n)) : Q] =ϕ(n)

2.

Proof. Recall thatGal(Q(ζn)/Q) ∼= Z/n×,

where the residue class of r acts by sending ζn to ζrn. Complex conjugation corresponds to the

residue class of −1 ≡ n− 1 mod n. Recalling that eθi = cos θ + sin θ i, and

cos θ =12(eθi + e−θi),

we obtain

cos(2π/n) =12(ζn + ζn).

Complex conjugation fixes each of the real numbers cos(2πk/n) for k = 1, 2, . . . , n − 1. Theresidue class of r acts by sending cos(2π/n) to cos(2πr/n); it is elementary to show thatcos(2πr/n) 6= cos(2π/n) unless r ≡ 1 mod n. Hence

〈( )〉 = {id, ( )} = Gal(Q(cos(2π/n))/Q).

Thus we haveQ(ζn)〈( )〉 = Q(cos(2π/n)),

and so [Q(cos(2π/n)) : Q] = ϕ(n)/2. ¤

Example 6.4. We have[Q(ζ24) : Q] = ϕ(24) = 8

andGal(Q(ζ24)/Q) = Z/2× Z/2× Z/2.


Proof. By Theorem 1.38 we have [Q(ζ24) : Q] = 8. Also,

ζ624 = i, ζ3

24 =12

+√

32

i, ζ824 = −1

2+√

32

i,

and all of these numbers are in Q(ζ24), hence Q(√

2,√

3, i) 6 Q(ζ24). It is easy to check that

[Q(√

2,√

3, i) : Q] = 8,

which implies thatQ(ζ24) = Q(

√2,√

3, i).

Using this we find thatGal(Q(ζ24)/Q) ∼= Z/2× Z/2× Z/2.

We also have cos(2π/24) = cos(π/12) ∈ Q(ζ24). Since

cos(2π/12) = cos(π/6) =√

32

,

we have

2 cos2(π/12)− 1 =√

32

and so4 cos4(π/12)− 4 cos2(π/12) + 1 =

34,

giving16 cos4(π/12)− 16 cos2(π/12) + 1 = 0.

Then16X4 − 16X2 + 1 = 16minpolyQ,cos(π/12)(X).

Note that case (i) of Kaplansky’s Theorem 4.20 applies to the polynomial minpolyQ,cos(π/12)(X).For this example, Gal(Q(ζ24)/Q) has 23 − 1 = 7 subgroups of each of the orders 2 and 4; it

is an interesting exercise to find them together with their fixed subfields. ¤

Remark 6.5. The minimal polynomial for cos(π/12) can also be found as follows. We haveΦ24(ζ24) = 0, hence since

Φ24(X) = X8 −X4 + 1,

we obtainζ824 − ζ4

24 + 1 = 0.

Then after multiplying by ζ−424 we have

ζ424 − 1 + ζ−4

24 = 0,

giving(ζ4

24 + ζ−424 )− 1 = 0.

Now(ζ24 + ζ−1

24 )4 = (ζ424 + ζ−4

24 ) + 4(ζ224 + ζ−2

24 ) + 6,

henceζ424 + ζ−4

24 = (ζ24 + ζ−124 )4 − 4(ζ2

24 + ζ−224 )− 6.

Similarly,(ζ24 + ζ−1

24 )2 = ζ224 + ζ−2

24 + 2,

soζ224 + ζ−2

24 = (ζ24 + ζ−124 )2 − 2.

Combining these we have

(ζ24 + ζ−124 )4 − 4(ζ24 + ζ−1

24 )2 + 1 = 0,

6.3. ARTIN’S THEOREM ON LINEAR INDEPENDENCE OF CHARACTERS 71

and so

16 cos4(π/12)− 16 cos2(π/12) + 1 = 0.

This method will work for any n where ϕ(n) is even, and this occurs whenever n > 2.

Now let K be a field with characteristic charK - n. The polynomial Φn(X) has integercoefficients, so we can view it as an element of K[X] since either Q 6 K or Fp 6 K and wecan reduce the coefficients modulo p. In either case it can happen that Φn(X) factors in K[X].However, we can still describe the splitting field of Xn − 1 over K and its Galois group.

Theorem 6.6. If charK - n, then the splitting field of Xn−1 over K is K(ζ), where ζ ∈ K

is a primitive n-th root of unity and Gal(K(ζ)/K) is abelian with order dividing ϕ(n).

Proof. Working in K, we know that Φn(ζ) = 0, hence the roots of minpolyK,ζ(X) ∈ K[X]are primitive roots of 1. So Xn − 1 splits over K(ζ) and each element α ∈ Gal(K(ζ)/K) hasthe action α(ζ) = ζrα , where gcd(rα, n) = 1. Hence Gal(K(ζ)/K) is isomorphic to a subgroupof Gal(Q(ζn)/Q) ∼= (Z/n)×, in particular it is abelian. ¤

Remark 6.7. When p = charK > 0, this Galois group only depends on the largest subfieldof K which is algebraic over Fp. For example, if K = Fpd(T ) then the value of d is the crucialfactor. The precise outcome can be determined with the aid of Proposition 5.21.

Example 6.8. We have the following splitting fields and Galois groups.

i) The splitting field of X4 − 1 over F3(T ) is F9(T ) and

Gal(F9(T )/F3(T )) ∼= (Z/4)× ∼= Z/2.

ii) By Proposition 5.20, X4− 1 splits over F5(T ) and the Galois group Gal(F5(T )/F5(T ))is trivial.

Proof. (i) By Proposition 5.20, X4− 1 is separable over F3(T ) and has irreducible factors(X − 1), (X + 1) and (X2 + 1). The splitting field of (X2 + 1) over F3 is F9 = F3(ζ), whereζ2 + 1 = 0, so (X2 + 1) splits over F9(T ). Also,

Gal(F9/F3) ∼= (Z/4)× ∼= Z/2,

with generator σ satisfying σ(ζ) = ζ−1 = −ζ. This generator clearly extends to an automor-phism of F9(T ) which fixes T .

(ii) By Proposition 5.20, X4 − 1 splits over F5. ¤

6.3. Artin’s Theorem on linear independence of characters

Let G be a group and K a field.

Definition 6.9. A group homomorphism χ : G −→ K× is called a character of G withvalues in K.

Example 6.10. Given any ring homomorphism ϕ : R −→ K we obtain a character of R×

in K by restricting ϕ to a map χϕ : R× −→ K×.

Example 6.11. Given an automorphism α : K −→ K, χα : K× −→ K× is a character ofK× in K.

Example 6.12. Let E/K be a Galois extension and σ ∈ Gal(E/K). Then χσ : E× −→ E×

is a character.


Definition 6.13. Let χ1, . . . , χn be characters of a group G in a field K. Then χ1, . . . , χn

are linearly independent if for t1, . . . , tn ∈ K,

t1χ1 + · · ·+ tnχn = 0 =⇒ t1 = · · · = tn = 0.

If χ1, . . . , χn are not linearly independent they are linearly dependent.

In this definition, the functional equation means that for all g ∈ G,

t1χ1(g) + · · ·+ tnχn(g) = 0.

Theorem 6.14 (Artin’s Theorem). Let χ1, . . . , χn be distinct characters of a group G in afield K. Then χ1, . . . , χn are linearly independent.

Proof. We proceed by induction on n. For n = 1 the result is easily verified. For theinductive assumption, suppose that it holds for any n 6 k.

Let χ1, . . . , χk+1 be a set of k + 1 distinct characters for which there are t1, . . . , tk+1 ∈ K

not all zero and such that

(6.1) t1χ1 + · · ·+ tk+1χk+1 = 0.

If one of the ti is zero, say tr = 0, then χ1, . . . , χr−1, χr+1, . . . , χk+1 is linearly dependent,contradicting the inductive assumption. Hence all of the ti must be non-zero. As χ1 6= χ2, theremust be an element g0 ∈ G for which χ1(g0) 6= χ2(g0). So for all g ∈ G, Equation (6.1) appliedto g0g yields

t1χ1(g0g) + · · ·+ tk+1χk+1(g0g) = 0,

and therefore since χj(g0g) = χj(g0)χj(g), we see that

t1χ1(g0)χ1 + · · ·+ tk+1χk+1(g0)χk+1 = 0.

Multiplying Equation (6.1) by χ1(g0) and subtracting gives

t2(χ2(g0)− χ1(g0))χ2 + t3(χ3(g0)− χ1(g0))χ3 + · · ·+ tk+1χk+1 = 0,

in which the coefficient t2(χ2(g0)−χ1(g0)) is not zero. Hence χ2, . . . , χk+1 is linearly dependent,again contradicting the inductive assumption. So χ1, . . . , χk+1 is linearly independent, whichdemonstrates the inductive step. ¤

Suppose that E/K is a finite Galois extension with cyclic Galois group Gal(E/K) = 〈σ〉 oforder n. For each u ∈ E×, the element uσ(u) · · ·σn−1(u) ∈ E satisfies

σ(uσ(u) · · ·σn−1(u)) = σ(u) · · ·σn−1(u)σn(u) = σ(u) · · ·σn−1(u)u,

hence in uσ(u) · · ·σn−1(u) ∈ E〈σ〉 = K. Using this we can define a group homomorphism

NE/K : E× −→ K×; NE/K(u) = uσ(u) · · ·σn−1(u).

NE/K is called the norm mapping for E/K and it generalizes the norm mapping for finite fieldsof Section 5.3.

There is another homomorphism

δE/K : E× −→ E×; δE/K(u) = uσ(u)−1.

Notice that for u ∈ E×,

NE/K(δE/K(u)) = (uσ(u)−1)(σ(u)σ2(u)−1 · · ·σn−1(u)σn(u)−1) = 1,

since σn(u) = u. So im δE/K 6 kerNE/K . Our next result is an important generalization ofProposition 5.27.

Theorem 6.15 (Hilbert’s Theorem 90). Let E/K be a finite Galois extension with cyclicGalois group Gal(E/K) = 〈σ〉 of order n. Then im δE/K = kerNE/K . Explicitly, if u ∈ E× anduσ(u) · · ·σn−1(u) = 1, then there is a v ∈ E× such that u = vσ(v)−1.

6.4. RADICAL EXTENSIONS 73

Proof. Let u ∈ kerNE/K .The characters σk : E× −→ E× with k = 0, 1, . . . , n−1 are distinct and linearly independent

by Artin’s Theorem 6.14. Consider the function

id +uσ + uσ(u)σ2 + · · ·+ uσ(u) · · ·σn−2(u)σn−1 : E× −→ E.

This cannot be identically zero, so for some w ∈ E, the element

v = w + uσ(w) + uσ(u)σ2(w) + · · ·+ uσ(u) · · ·σn−2(u)σn−1(w)

is non-zero. Notice that

uσ(v) = uσ(w) + uσ(u)σ2(w) + uσ(u)σ2(u)σ3(w) + · · ·+ uσ(u)σ2(u) · · ·σn−1(u)σn(w) = v,

sinceuσ(u)σ2(u) · · ·σn−1(u)σn(w) = w.

Thus we have u = vσ(v)−1 as required. ¤

6.4. Radical extensions

In this section we will investigate splitting fields of polynomials of the form Xn − a, wherecharK - n.

Proposition 6.16. Let f(X) = Xn − a ∈ K[X] be irreducible and separable over K. Thenthe splitting field of f(X) over K has the form E = K(u, ζ), where u is a root of f(X) and ζ

is a primitive n-th root of unity.

Corollary 6.17. If K contains a primitive n-th root of unity ζ then the splitting field off(X) = Xn − a over K has the form E = K(u), where u is a root of f(X). The Galois groupGal(K(f(X))/K) is cyclic of order n with a generator σ for which σ(u) = ζu.

For the more general situation of Proposition 6.16 we have

{id} / Gal(K(ζ)/K) / Gal(K(f(X))/K(ζ)),

where Gal(K(ζ)/K) and Gal(K(ζ)/K))/ Gal(K(ζ, u)/K(ζ) are abelian and in fact cyclic. TheGalois Correspondence identifies the following towers of subfields and subgroups.

K(ζ, u)``

''

Gal(K(ζ, u)/K)

/

88

~~

K(ζ)ee 99

Gal(K(ζ)/K)

K {id}

Definition 6.18. Let K be a field of characteristic not dividing n and which contains aprimitive n-th root of 1, ζ say. Then L/K is a simple n-Kummer extension if L = K(u) whereun = a for some a ∈ K. L/K is an (iterated) n-Kummer extension if L = K(u1, . . . , uk) whereun

1 = a1, ..., unk = ak for some elements a1, . . . , ak ∈ K.

Note that we do not require the polynomials Xn − aj ∈ K[X] to be irreducible in thisdefinition.

Proposition 6.19. Let K(u)/K be a simple n-Kummer extension. Then K(u)/K is aGalois extension and Gal(K(u)/K) is cyclic with order dividing n.


Proof. Suppose that un = a ∈ K. Then in K[X] we have

Xn − a = (X − u)(X − ζu) · · · (X − ζn−1u).

Clearly the roots of Xn − a are distinct and so K(u)/K is separable over K; in fact, K(u) is asplitting field of Xn − a over K. This means that K(u)/K is Galois.

For each α ∈ Gal(K(u)/K) we have α(u) = ζrαu for some rα = 0, 1 . . . , n− 1. Notice thatfor β ∈ Gal(K(u)/K),

βα(u) = β(ζrαu) = ζrαβ(u) = ζrαζrβu = ζrα+rβu,

and so rβα = rα + rβ. Hence the function

ρ : Gal(K(u)/K) −→ 〈ζ〉 ; ρ(α) = ζrα ,

is a group homomorphism. As 〈ζ〉 is cyclic of order n, Lagrange’s Theorem implies that theimage of ρ has order dividing n. Since every element of Gal(K(u)/K) is determined by its effecton u, ρ is injective, hence |Gal(K(u)/K)| divides n. In fact, Gal(K(u)/K) is cyclic since everysubgroup of a cyclic group is cyclic. ¤

Example 6.20. Let n > 1 and q ∈ Q. Then Q(ζn, n√

q)/Q(ζn) is a simple n-Kummerextension.

Example 6.21. Q(i,√

2)/Q(i) is a simple 4-Kummer extension for which Gal(Q(i,√

2)/Q(i))is cyclic of order 2.

Proof. We have (√

2)4 − 4 = 0, but

X4 − 4 = (X2 − 2)(X2 + 2),

and

X2 − 2 = minpolyQ(i),√

2(X).

The corresponding group homomorphism ρ : Gal(Q(i)(√

2)/Q(i)) −→ 〈i〉 has image

im ρ = {1,−1} 6 〈i〉 . ¤

Here is a converse to Proposition 6.19.

Proposition 6.22. Suppose that charK - n and K contains a primitive n-th root of unity,ζ say. If E/K is a finite Galois extensions with cyclic Galois group of order n, then there isan element a ∈ E such that E = K(a) and a is a root of a polynomial of the form Xn − b withb ∈ K. Hence E/K is a simple n-Kummer extension.

Proof. We haveNE/K(ζ) = ζn = 1,

so by Hilbert’s Theorem 6.15, there is an element a ∈ E for which ζ = aσ(a)−1. Then σ(a) =ζ−1a and the elements σk(a) = ζ−ka for k = 0, 1, . . . , n − 1 are distinct and so must be the n

conjugates of a. Note also that

Xn − an = (X − a)(X − ζa) · · · (X − ζn−1a) = (X − a)(X − σ(a)) · · · (X − σn−1(a)),

so an ∈ K since it is fixed by σ. This shows that K(a) 6 E and

n = [K(a) : K] 6 [E : K] = n,

therefore [K(a) : K] 6 [E : K] = n and K(a) = E. ¤

6.5. SOLVABILITY AND RADICAL EXTENSIONS 75

6.5. Solvability and radical extensions

We begin by recalling some ideas about groups, see [2, 3] for further details.

Definition 6.23. A group G is solvable, soluble or soluable if there is a chain of subgroups(called a subnormal series)

{1} = G` 6 G`−1 6 · · · 6 G1 6 G0 = G

in which Gk+1 / Gk and each composition factor Gk/Gk+1 is abelian; we usually write

{1} = G` / G`−1 / · · · / G1 / G0 = G.

If each composition factor is a cyclic group of prime order the subnormal series is called acomposition series. A group which is not solvable is called insolvable.

Remark 6.24. For a solvable group, it is a standard result that we can always refine asubnormal series (i.e., add extra terms) to obtain a composition series. The primes as well asthe number of times each occurs are all determined by |G|, only the order of these varying fordifferent composition series.

Example 6.25. Let G be a finite abelian group. Then G is solvable.

Example 6.26. Let G be a finite p-group, where p is a prime. Then G is solvable.

In fact, for a finite p-group G, there is always a normal subgroup of a p-group with index p,so in this case we can assume each quotient Gk/Gk+1 is cyclic of order p.

Proposition 6.27. Let G be a group.

i) If G is solvable then every subgroup H 6 G and every quotient group G/N is solvable.ii) If N / G and G/N are solvable then so is G.

In the opposite direction we can sometimes see that a group is insolvable. Recall that agroup is simple if it has no non-trivial proper normal subgroups.

Proposition 6.28. Let G be a finite group. Then G is insolvable if any of the followingconditions holds:

i) G contains a subgroup which is a non-abelian simple group (or has a quotient groupwhich is a non-abelian simple group).

ii) G has a quotient group which is a non-abelian simple group.iii) G has a composition series in which one of the terms is a non-abelian simple group.

Example 6.29. For n > 5, the alternating and symmetric groups An and Sn are insolvable.

Proof. This follows from the fact that if n > 5, An is a simple group and An / Sn withquotient group Sn/An

∼= Z/2. ¤

Now we explain how this relates to fields and their extensions. Let K be a field and L/K afinite extension. For simplicity, we assume also that charK = 0.

Definition 6.30. L/K is a radical extension of K if it has the form L = K(a1, a2, . . . , an)with

adkk ∈ K(a1, a2, . . . , ak−1)

for some dk > 1. Thus every element of L is expressible in terms of iterated roots of elementsof K.

Definition 6.31. If L is the splitting field of a polynomial f(X) ∈ K[X], then f(X) issolvable by radicals over K if L is contained in a radical extension of K.


Definition 6.32. L/K is solvable if L 6 L′ where L′/K is a finite radical Galois extensionof K.

Theorem 6.33. Let E/K be a finite Galois extension. Then E/K is solvable if and only ifthe group Gal(E/K) is solvable.

Proof. Suppose that E 6 E′ where E′/K is a radical Galois extension. Then Gal(E/K)is a quotient group of Gal(E′/K), so it is solvable by Proposition 6.27.

Now suppose that Gal(E/K) is solvable and let n = |Gal(E/K)|. Let E′ be the splittingfield of Xn − 1 over E, so E′ contains a primitive n-th root of unity ζ and therefore it containsa primitive d-th root of unity for every divisor d of n. Now Gal(E′/E) / Gal(E′/K) andby Theorem 6.6, Gal(E′/E) is abelian. Also, Gal(E′/K)/Gal(E′/E) ∼= Gal(E/K) which issolvable, so Gal(E′/K) is solvable by Proposition 6.27. We will now show that E′/K is aradical extension.

Clearly K(ζ)/K is radical. Then Gal(E′/K(ζ)) / Gal(E′/K) is solvable. Let

{1} = G` / G`−1 / · · · / G1 / G0 = Gal(E′/K(ζ))

be a composition series. The extension (E′)G1/K(ζ) is radical by Proposition 6.22. Similarly,each extension (E′)Gk+1/(E′)Gk is radical. Hence E′/K(ζ) is radical, as is E′/K. ¤

Example 6.34. The Galois group of the extension Q(ζ3,3√

2)/Q is solvable.

Proof. We have already studied this extension in Example 3.30 and 4.19. ClearlyQ(ζ3,3√

2)is a radical extension of Q and

Q(ζ3,3√

2) = Q(ζ3)(3√

2).

We know that Gal(Q(ζ3,3√

2)/Q) ∼= S3, where we identify each element of the Galois group witha permutation of the three roots of X3 − 2 in Q(ζ3,

3√

2) which we list in the order3√

2,3√

2 ζ3,3√

2 ζ23 .,

We have the following towers of subfields and subgroups related under the Galois Correspon-dence.

Q(ζ3,3√

2)ee

))

S355

yy

Q(ζ3) = Q(ζ3,3√

2)A3

3

ii 55A3 = Gal(Q(ζ3,

3√

2)/Q(ζ3))

2

Q

2

{id}3

Here Q(ζ3)/Q is itself a Galois extension and A3 / S3. Notice that A3∼= Z/3 and S3/A3

∼= Z/2,so we have a composition series for S3,

{id} / A3 / S3. ¤

It is also interesting to reverse the question and ask whether there are extensions which arenot solvable. This was a famous problem pursued for several hundred years. To find examples,we first recall that the smallest non-abelian simple group is A5 which has order 60. We shouldtherefore expect to look for a polynomial of degree at least 5 to find a Galois group for a splittingfield to be simple or occur as a composition factor of such a Galois group. Here is an explicitexample over Q.

Example 6.35. The splitting field of the polynomial f(X) = X5 − 35X4 + 7 ∈ Q[X] is notsolvable by radicals.

6.6. SYMMETRIC FUNCTIONS 77

Proof. Let E = Q(f(X)) be the splitting field of f(X) over Q. Using the EisensteinCriterion 1.34 with p = 7, we find that f(X) is irreducible over Q. By Theorem 4.7(iii),5 divides the order of Gal(E/Q), so this group contains an element of order 5 by Cauchy’sLemma.

Now observe that

f ′(X) = 5X4 − 140X3 = 5X3(X − 28), f ′′(X) = 20X4 − 420X2 = 20X2(X − 21).

There are two turning points, namely a maximum at x = 0 and a minimum at x = 28. Then

f(0) = 7 > 0 > f(28) = −4302585,

hence there are three real roots of f(X) and two non-real complex ones. Then complex conju-gation restricts to an element of order 2 in Gal(E/Q) which interchanges the non-real roots andfixes the others. If we list the roots of f(X) as u1, u2, u3, u4, u5 with u1, u2 being the no-realroots, then the transposition (1 2) ∈ S5 corresponds to this element. Furthermore, the onlyelements of S5 of order 5 are 5-cycles; by taking an appropriate power we can assume that thereis a 5-cycle of the form (1 2 3 4 5) corresponding to an element of Gal(E/Q) which we can viewas a subgroup of S5. The next lemma shows that Gal(E/Q) ∼= S5.

Lemma 6.36. Let n > 1. Suppose that H 6 Sn and H contains the elements (1 2) and(1 2 · · · n). Then H = Sn.

The proof is left as an exercise. This completes the verification of the Example. ¤

It is worth remarking that the most extreme version of this occurs when we ask for a Galoisgroup which is simple. There has been a great deal of research activity on this question in thepast few decades, but apparently not all simple groups are known to occur as Galois groups ofextensions of Q or other finite subextensions of C/Q. Here is an example whose Galois groupis A5; this is verified using Proposition 6.44.

Example 6.37. The Galois group of f(X) = X5+20X+16 over Q is Gal(Q(f(X))/Q) ∼= A5.

6.6. Symmetric functions

Let k be a field. Consider the polynomial ring on n indeterminates k[X1, . . . , Xn] and itsfield of fractions K = k(X1, . . . , Xn). Each permutation σ ∈ Sn acts on k[X1, . . . , Xn] by

σ · f(X1, . . . , Xn) = fσ(X1, . . . , Xn) = f(Xσ(1), . . . , Xσ(n)).

Viewed as a function σ· : k[X1, . . . , Xn] −→ k[X1, . . . , Xn] is a ring isomorphism; this extends toa ring isomorphism σ· : k(X1, . . . , Xn) −→ k(X1, . . . , Xn). Varying σ we obtain actions of thegroup Sn on k[X1, . . . , Xn] and k(X1, . . . , Xn) by ring isomorphisms fixing k and in the lattercase it is by field automorphisms fixing k.

Definition 6.38. The field of symmetric functions on n indeterminates is

Symn(k) = k(X1, . . . , Xn)Sn 6 k(X1, . . . , Xn).

So if f(X1, . . . , Xn) ∈ k(X1, . . . , Xn), then

f(X1, . . . , Xn) ∈ Symn(k) ⇐⇒ ∀σ ∈ Sn f(X1, . . . , Xn) = f(Xσ(1), . . . , Xσ(n)).

Theorem 6.39. The extension k(X1, . . . , Xn)/Symn(k) is a finite Galois extension forwhich Gal(k(X1, . . . , Xn)/Symn(k)) ∼= Sn.


Proof. There are elements of k[X1, . . . , Xn] ⊆ k(X1, . . . , Xn) called elementary symmetricfunctions,

ek =∑

i1<i2<···<ik

Xi1Xi2 · · ·Xik ,

where 1 6 k 6 n. It is easy to see that for every σ ∈ Sn, eσk = ek, so ek ∈ Symn(k). Working in

the ring k(X1, . . . , Xn)[Y ] we have

fn(Y ) = Y n − e1Yn−1 + · · ·+ (−1)n−1en−1Y + (−1)nen = 0,

hence the roots of this polynomial are the Xi. So k(X1, . . . , Xn) is the splitting field of fn(Y )over Symn(k). Now Sn 6 Gal(k(X1, . . . , Xn)/Symn(k)), hence

[k(X1, . . . , Xn) : Symn(k)] = |Gal(k(X1, . . . , Xn)/ Symn(k))| > |Sn| = n!.

But as every element of Gal(k(X1, . . . , Xn)/Symn(k)) permutes the roots of fn(Y ) and is de-termined by this permutation, we also have

n! > |Gal(k(X1, . . . , Xn)/Symn(k))|.Combining these inequalities we obtain |Gal(k(X1, . . . , Xn)/Symn(k))| = n! and thereforeGal(k(X1, . . . , Xn)/Symn(k)) = Sn. ¤

Remark 6.40. In fact, this proof shows that the extension k(X1, . . . , Xn)/k(e1, . . . , en) isGalois of degree n!. Since k(e1, . . . , en) 6 Symn(k) we can also deduce that k(e1, . . . , en) =Symn(k). Hence every element of Symn(k) is a rational function in the ei. Analogous resultsare true for polynomials, i.e.,

k[X1, . . . , Xn]Sn = k[e1, . . . , en].

Corollary 6.41. If n > 5, the extension k(X1, . . . , Xn)/Symn(k) is not solvable.

6.7. Galois groups of even and odd permutations

We have seen that for a monic separable polynomial f(X) ∈ K[X] of degree n, the Galoisgroup of its splitting field Ef = K(f(X)) can naturally be thought of as a subgroup of thesymmetric group Sn, where we view the latter as permuting the roots of f(X). It is reasonableto ask when Gal(Ef/K) 6 An rather than just Gal(Ef/K) 6 Sn.

We first recall an interpretation of the sign of a permutation σ ∈ Sn, sgnσ = ±1. For eachpair i, j with 1 6 i < j 6 n, exactly one of the inequalities σ(i) < σ(j) or σ(j) < σ(i) musthold and the ratio (σ(j) − σ(i))/(j − i) is either positive or negative. It is easily verified thatthe right-hand side of the following equation must have value ±1 and we have

(6.2) sgnσ =∏

16i<j6n

σ(j)− σ(i)j − i

.

Note that this is sometimes used as the definition of sgnσ.Suppose that f(X) factorizes over Ef as

f(X) = (X − u1) · · · (X − un) =n∏

i=1

(X − ui).

Here u1, . . . , un ∈ Ef are the roots of f(X); as we have assumed that f(X) is separable, the ui

are distinct.

Definition 6.42. The discriminant of f(X) is

∆(f(X)) =∏

16i<j6n

(uj − ui)2 ∈ Ef .

Notice that ∆(f(X)) 6= 0 since ui 6= uj if i 6= j.

6.7. GALOIS GROUPS OF EVEN AND ODD PERMUTATIONS 79

Proposition 6.43. For every σ ∈ Gal(Ef/K), σ(∆(f(X))) = ∆(f). Hence ∆(f(X)) ∈ K.

Proof. For σ ∈ Gal(Ef/K) 6 Sn, we have

σ(∆(f(X))) =∏

16i<j6n

(uσ(j) − uσ(i))2 =

∏

16i<j6n

(uσ(j) − uσ(i))

2

.

Now for each pair i, j with i < j,

σ(uj − ui) = uσ(j) − uσ(i) = ±(uj − ui),

so

(6.3)∏

16i<j6n

(uσ(j) − uσ(i)) = sgnσ∏

16i<j6n

(uj − ui) = (±1)∏

16i<j6n

(uj − ui).

Hence σ(∆(f(X))) = ∆(f(X)). Since EGal(Ef /K)f = K, this implies that ∆(f(X)) ∈ K. ¤

There are explicit formulæ for expressing ∆(f) in terms of the coefficients of the polynomial

f(X) = a0 + a1X + · · ·+ an−1Xn−1 + Xn.

Here are the first few examples obtained using Maple.

∆(a0 + a1X + X2) = −4a0 + a21.n = 2:

∆(a0 + a1X + a2X2 + X3) = −27a2

0 + 18a0a1a2 + a21a

22 − 4a3

2a0 − 4a31.n = 3:

n = 4: ∆(a0 + a1X + a2X2 + a3X

3 + X4) = 18a3a31a2 − 6a2

3a21a0 − 192a3a1a

20 − 27a4

1

+ 144a2a23a

20 + 144a0a

21a2 + 256a3

0 − 4a33a

31 − 128a2

2a20 + 16a4

2a0 − 4a32a

21

+ 18a33a1a2a0 − 80a3a1a

22a0 − 27a4

3a20 + a2

2a23a

21 − 4a3

2a23a0.

n = 5: ∆(a0 + a1X + a2X2 + a3X

3 + a4X4 + X5) = 2250a4a

23a

30 − 36a0a

34a

31 − 128a2

3a41

+ 2000a20a3a

21 − 900a1a

33a

20 − 2500a3

0a4a1 − 50a20a

24a

21 − 900a4a

32a

20 − 27a4

4a41 − 3750a3a2a

30

+ 356a32a2

2a4a1a0 + 560a3a22a

24a

20 − 2050a3a2a

20a4a1 − 80a2

3a2a4a13 − 630a3

3a2a4a02

+ 825a23a

22a

20 + 16a3

3a32a0 + 2000a2a

24a

30 − 6a2

2a24a

31 − 128a2

2a44a

20 + 16a4

2a34a0 − 4a3

2a34a

21

+ 108a53a

20 + 108a5

2a0 − 746a3a2a0a42a1

2 − 27a42a

21 + 256a5

4a30 − 4a3

3a22a

21 + 144a3a

22a

31

+ 144a24a

41a3 + 3125a4

0 + 256a51 − 72a4

3a2a1a0 + 18a3a2a34a

31 + 560a2

3a2a0a21 + 16a4

3a31

+ 18a3a23a4a1

2 − 72a3a24a4a0 + 144a3

2a2a43a0

2 − 192a44a1a3a0

2 − 630a3a23a1a0

+ 24a23a4

2a1a0 + a32a2

2a42a1

2 − 6a43a1

2a32a0 − 80a3a2

2a43a1a0 − 4a3

2a23a4

2a0

+ 2250a1a22a

20 − 1600a3a

34a

30 − 192a4a

41a2 − 1600a0a

31a2 − 4a3

3a31a

24 − 27a4

3a24a

20

+ 1020a42a3

2a02a1 + 18a3

3a2a42a0a1 + 160a2a4

3a02a1 + 144a2a4

4a0a12

+ 24a4a12a3

3a0 + 1020a0a4a22a1

2 + 160a0a4a13a3.

So for example,

∆(X5 + a4X4 + a0) = a3

0(3125a0 + 256a54), ∆(X5 + a1X + a0) = 256a5

1 + 3125a40.

Now let

δ(f(X)) =∏

16i<j6n

(uj − ui) ∈ Ef .


Then δ(f(X))2 = ∆(f), so the square roots of ∆(f(X)) are ±δ(f(X)). Now consider the effectof σ ∈ Gal(Ef/K) on δ(f(X)) ∈ Ef . By Equation (6.3),

σ(δ(f)) = sgnσ δ(f) = ±δ(f).

If δ(f(X)) ∈ K, this means that sgnσ = 1. On the other hand, if δ(f) /∈ K then

K(δ(f(X))) = EGal(Ef /K)∩An

f .

Of course |Gal(Ef/K)/Gal(Ef/K) ∩An| = 2.

Proposition 6.44. The Galois group Gal(Ef/K) 6 Sn is contained in An if and only if∆(f(X)) is a square in K. We summarize this in the form of a useful result.

For the polynomials of Examples 6.35 and 6.37 we have

∆(X5 − 35X4 + 7) = −4611833296875 = −33 · 56 · 74 · 29 · 157,

δ(X5 − 35X4 + 7) = ±53 · 3 · 72 ·√

3 · 29 · 157 i = ±18375√

13659 i /∈ Q;

∆(X5 + 20X + 16) = 1024000000 = 216 · 56,

δ(X5 + 20X + 16) = ±2853 ∈ Q.

81


6.1. Let p > 0 be a prime and G a group of order |G| = pn for some n > 1. Show by inductionon n that there is a normal subgroup N / G with |N | = pn−1. [Hint: what do you know aboutthe centre of G? Use this information to produce a quotient group of smaller order than G.]

6.2. Let K be a field for which charK 6= 2 and n > 1 be odd. If K contains a primitive n-throot of unity, show that then K contains a primitive 2n-th root of unity.

6.3. (a) Describe the elements of (Z/24)× explicitly and verify that this group is isomorphicto Z/2× Z/2× Z/2. Describe the effect of each element on Q(ζ24) and Q(cos(π/12) under theaction described in Theorem 6.2.(b) Determine the group (Z/20)× and describe the effect of each of its elements on Q(ζ20) andQ(cos(π/10) under the action described in Theorem 6.2.

6.4. Let n > 1.

(a) What can you say about sin(2π/n) and Gal(Q(sin(2π/n))/Q))?(b) Determine sin(π/12) and Gal(Q(sin(π/12))/Q)).

6.5. Find all values of n > 1 for which ϕ(n) | 4. Using this, determine which roots of unity liein the following fields:

Q(i), Q(√

2), Q(√

2 i), Q(√

3), Q(√

3 i), Q(√

5), Q(√

5 i).

6.6. Prove Lemma 6.36. [Hint: show that every 2-cycle of the form (i i + 1) is in H byconsidering elements of the form (1 2 · · · n)r(1 2)(1 2 · · · n)n−r.]

6.7. For n > 1 and 1 6 k 6 n, the k-th power sum sk ∈ k[X1, . . . , Xn]Sn is defined to be

sk =∑

16i6n

Xki .

Prove the formula

sk = e1sk−1 − e2sk−2 + · · ·+ (−1)k−1ek−1s1 + (−1)kkek.

6.8. For n > 1 and 1 6 k 6 n, the total symmetric function hk ∈ k[X1, . . . , Xn]Sn is defined tobe

hk =∑

j16j26···6jk

Xj1Xj2 · · ·Xjk,

i.e., the sum of all the monomials in the Xi of degree k.

(a) For large values of n, express h1, h2, h3 in terms of the three elementary symmetricfunctions e1, e2, e3.

(b) Show that the power sum functions sk of the previous question satisfy

sk = −(h1sk−1 + h2sk−2 + · · ·+ hk−1s1) + khk.

6.9. Let E/K be a Galois extension with cyclic Galois group Gal(E/K) = 〈σ〉 of order n.(a) Show that the function

E −→ E; u 7−→ u + σ(u) + σ2(u) + · · ·+ σn−1(u),


takes values in K and use this to define a K-linear mapping TrE/K : E −→ K.(b) If v ∈ E has TrE/K(v) = 0, show that there is a w ∈ E such that v = w − σ(w).[Hint: Show that there is an element t ∈ E for which TrE/K t 6= 0, then consider

w =1

(TrE/K t)(vσ(t) + (v + σ(v))σ2(t) + · · ·+ (v + σ(v)σ2(t) + · · ·+ σn−2(v))σn−1(t)

)

and adapt the proof of Hilbert’s Theorem 90 6.15, using TrE/K in place of NE/K .]

Solutions

Solutions for exercises on Chapter 1

1.1. (a) Let u, v ∈ S and suppose that uv = 0; then u = 0 or v = 0 since u, v ∈ R and R is anintegral domain. Consider the unit homomorphisms η : Z −→ R and η′ : Z −→ S. Then for alln ∈ Z, η′(n) = η(n), so ker η′ = ker η and therefore charS = charR.(b) Q is a field and Z ⊆ Q is a subring which is not a field.

1.2. (a) For any subring R ⊆ C, R is an integral domain with characteristic subring Z andcharR = 0.(b) The characteristic subring of A[X] is the same as that of A and charA[X] = charA. A[X]is an integral domain if and only if A is an integral domain.(c) If we identify A with the subring of scalar matrices in Matn(A), then the characteristicsubring of Matn(A) is the same as that of A and charMatn(A) = charA. If n > 1 thenMatn(A) is not commutative, in any case it always has zero-divisors since any singular matrixis a zero-divisor.

1.3. The main thing to check is that ϕ(u + v) = ϕ(u) + ϕ(v) which is a consequence of theIdiot’s Binomial Theorem. For R = Fp[X], ϕ is not surjective, while for R = Fp[X]/(X2), ϕ isnot injective.

1.4. (a) Recall from the Isomorphism Theorems of basic Ring Theory that ϕ−1Q / R; we needto show it is a prime ideal. Suppose that u, v ∈ R with uv ∈ ϕ−1Q; then ϕ(u)ϕ(v) = ϕ(uv) ∈ Q

and so ϕ(u) ∈ Q or ϕ(v) ∈ Q, hence u ∈ ϕ−1Q or v ∈ ϕ−1Q.(b) Consider the inclusion function inc : R −→ S; then inc−1 Q = Q ∩ R, so this result followsfrom (a).(c) Consider Z ⊆ Q; then the zero-ideal (0)Q /Q has (0)Q∩Z = (0)Z /Z but this is not maximalin Z since for any prime p > 0, (p)Z / Z is a (maximal) ideal that properly contains (0)Z.(d) We have P ⊆ Q ∩ R / R with P / R maximal; so P ⊆ Q ∩ R. In fact Q only needs to be aproper ideal of S for this argument to work.

1.5. The only proper ideal of k is the zero ideal (0), so kerϕ = (0).

1.6. (a) Addition and multiplication are given by the obvious formulæ,

(u1+v1i)+(u2+v2i) = (u1+u2)+(v1+v2)i, (u1+v1i)(u2+v2i) = (u1u2−v1v2)+(u1v2+u2v1)i,

with Z[i] and Q[i] both closed under these operations and containing 1 = 1 + 0i as a unity, sothey are subrings of the field C; by Qu. 1.1, they are both integral domains. To see that Q[i] isa field, notice that if u + vi 6= 0 with u, v ∈ Q,

(u− vi)(u + vi) = (u + vi)(u− vi) = u2 + v2 6= 0,

sou

u2 + v2+

v

u2 + v2i ∈ Q(i)

is the inverse of u + vi. Hence every non-zero element of Q[i] has an inverse, therefore Q[i] is afield.

84 SOLUTIONS

(b) & (c) The crucial point is that every element of Q[i] can be written as1n

(u + vi) withn, u, v ∈ Z and n 6= 0. Then

inc∗

((u + vi)

n

)= inc∗

((u + vi)n + 0i

)=

1n

(u + vi),

so the latter element is in the image of inc∗ which must therefore be a surjection.

1.7. (a) Addition and multiplication are given by the usual formulæ,

(∞∑

k=0

akXk) + (

∞∑

k=0

bkXk) =

∞∑

k=0

(ak + bk)Xk, (∞∑

k=0

akXk)(

∞∑

k=0

bkXk) =

∞∑

k=0

(k∑

`=0

a`bk−`)Xk.

Clearly k[X] ⊆ k[[X]] is a subring. Given two non-zero elements a, b ∈ k[[X]] we may write

a =∞∑

k=k0

akXk, b =

∞∑

`0

b`X`

with ak0 6= 0 6= b`0 . Then the lowest degree term in ab is ak0b`0Xk0+`0 with ak0b`0 6= 0. Hence

ab 6= 0. So k[[X]] is an integral domain.(b) Let a =

∑∞k=0 akX

k ∈ k[[X]]. Then a has an inverse in k[[X]] only if there is a b =∑∞k=0 b`X

` ∈ k[[X]] with ab = 1, in particular this forces a0 6= 0 since otherwise the lowest termin X in ab would be of degree greater than 0. Conversely, if a0 6= 0, then we can inductivelysolve the system of equations

a0b0 = 1,n∑

`=0

a`bn−` = a0bn + a1bn−1 + · · ·+ anbn = 0 (n > 1),

to ensure that ab = 1.(c) We can define make the set k((X)) of all such finite tailed Laurent series into a ring withaddition and multiplication defined by

(∞∑

k=k1

akXk) + (

∞∑

k=k2

bkXk) =

∞∑

k=min{k1,k2}(ak + bk)Xk,

(∞∑

k=k0

akXk)(

∞∑

`=`0

b`X`) =

∞∑

k=min{k0,`0}(

k∑

j=0

a`bk−j)Xk.

Clearly k[[X]] ⊆ k((X)) is a subring. Notice that every element∑∞

k=k0akX

k ∈ k((X)) withk0 < 0 can be written as

(∑

r=0

ar+k0Xr)Xk0 .

The inclusion inc : k[[X]] −→ k((X)) extends to the monomorphism inc∗ : Fr(k[[X]]) −→ k((X))for which

inc∗

(∑∞r=0 ar+k0X

r

X−k0

)= (

∞∑

r=0

ar+k0Xr)Xk0 ,

so inc∗ is surjective.

1.8. Here f(X) = (−9X + 7)d(X) + (3X − 3).

1.9. Here f(X) = −X3−X2+X+1 and d(X) = −X3−X with f(X) = d(X)+(−X−X2+1) =d(X) + (2X2 + 2X + 1).

1.10. The reduction modulo p function

ρ : Z[X] −→ Fp[X]; ρ(f(X)) = f(X),

85

is a ring homomorphism. If f(X) = g(X)h(X) with g(X), h(X) ∈ Z[X], deg g(X) < deg f(X)and deg h(X) < deg f(X), then

f(X) = ρ(g(X)h(X)) = ρ(g(X))ρ(h(X)) = g(X)h(X),

where deg g(X) < deg f(X) = deg f(X) and deg h(X) < deg f(X) = deg f(X). But this isimpossible since f(X) is irreducible. So f(X) must be irreducible.X3 −X + 1 reduces modulo 3 to an irreducible since it has no roots modulo 3. So X3 −X + 1is irreducible.X3 + 2X + 1 ≡ X3 −X + 1 mod 3 so this polynomial reduces modulo 3 to an irreducible andso is irreducible.X3 + X − 1 reduces modulo 2 to an irreducible since it has no roots modulo 2. So X3 + X − 1is irreducible.X5 −X + 1 is irreducible modulo 3 and 5 so is itself irreducible.X5 + X − 1 = (X3 + X2 − 1)(X2 −X + 1) and 5X3 − 10X + X2 − 2 = (5X + 1)(X2 − 2) soneither of these is irreducible.

1.11. I1 = (X2+1), I2 = (X2+2), I3 = (X2−2), I4 = (X−√2), I5 = (X2+2), I6 = X2+X+1.

1.12. The image isε√2Q[X] = Q[

√2] = {a + b

√2 : a, b ∈ Q}.

The image of ε−√2 is ε−√2Q[X] = Q[√

2] = ε√2Q[X]. We have

ker ε√2 = ker ε−√2 = (X2 − 2) /Q[X]

which is a maximal ideal.

1.13. Notice that ω = (−1 +√

3i)/2 = ζ3 is a primitive 3-rd root of unity and is a root of theirreducible polynomial X2 + X + 1 ∈ Q[X]. Then

εω Q[X] = Q[ω] = {a + bω : a, b ∈ Q}, ker εω = (X2 + X + 1) /Q[X],

where (X2 + X + 1) /Q[X] is a maximal ideal. The other complex root of X2 + X + 1is ω2 sothe evaluation homomorphism εω2 has εω2 Q[X] = εω Q[X] and ker εω2 = ker εω.

1.14. We have

εαQ[X] = Q[α] = {a + bα + cα2 + dα3 : a, b, c, d ∈ Q}, ker εα = (X4 − 2) /Q[X],

and the latter ideal is maximal. The other complex roots of X4 − 2 are −α, α i,−α i (noticethat two of these are real while the other two are not). Then

ker ε−α = ker εα i = ker ε−α i = (X4 − 2) /Q[X]

but although ε−αQ[X] = Q[α], we have

εα iQ[X] = ε−α iQ[X] = Q[α i] = {a + bα i + cα2 + dα3 i : a, b, c, d ∈ Q} 6= Q[α],

so εα iQ[X] 6= εαQ[X] since one of these is a subset of R but the other is not.If we replace Q by R, then X4 − 2 = (X2 − 2)(X2 + 2) in R[X] and we have to consider

which factor α is a root of. If α2 − 2 = 0 then

εαR[X] = ε−αR[X] = R[α] = {a + bα : a, b ∈ R} ⊆ R, ker εα = ker ε−α = (X2 − 2) / R[X].

If α2 + 2 = 0 then

εαR[X] = ε−αR[X] = R[α] = {a + bα : a, b ∈ R} 6⊆ R, ker εα = ker ε−α = (X2 + 2) / R[X].

1.15. To see that the homomorphism

Aff1(k) −→ Aut|(k[X]); A 7−→ αA−1 ,

86 SOLUTIONS

described in the Proof of Example 1.52 is surjective, suppose that ϕ ∈ Aut|(k[X]) is anyautomorphism. Let

ϕ(X) = a0 + a1 + · · ·+ anXn

with ai ∈ k and an 6= 0. If n = 0 then ϕk[X] = k ⊆ k[X] so ϕ would not be surjective, hencewe must have n > 1. Suppose that show that n > 1. Then

ϕk[X] = {c + 0 + c1ϕ(X) + · · ·+ ckϕ(X)k : c0, c1, . . . , ck ∈ k} = k[X].

But if k > 0 and ck 6= 0 then deg(c + 0 + c1ϕ(X) + · · · + ckϕ(X)k) = kn > 1, so X /∈ ϕk[X],which gives a contradiction. So we must have n = 1. Therefore ϕ(X) = a0+a1X and so ϕ = αA

for some A ∈ Aff1(k).

1.16. Calculation.

1.17. Heredeg Φ20(X) = ϕ(20) = ϕ(4)ϕ(5) = 2× 4 = 8

and

X20 − 1 = (X10 − 1)(X10 + 1) = (X10 − 1)(X2 + 1)(X8 −X6 + X4 −X2 + 1).

Since the roots of X10 − 1 are the 10-th roots of unity, we find that

Φ20(X) | (X2 + 1)(X8 −X6 + X4 −X2 + 1);

since cyclotomic polynomials are irreducible, we must have Φ20(X) = X8 −X6 + X4 −X2 + 1.


2.1. This is similar to Example 2.4.

2.2. It is obvious that [Q(√

p,√

q) : Q(√

p)] 6 2; if [Q(√

p,√

q) : Q(√

p)] = 1 then√

q ∈ Q(√

p),say

√q = a + b

√p for some a, b ∈ Q. Then

q = (a + b√

p)2 = (a2 + b2p) + 2ab√

p,

giving the simultaneous pair of equations

a2 + b2p = q, 2ab = 0.

If b = 0 then√

q ∈ Q which contradicts the result of Qu. 2.1. If a = 0 then√

q = b√

p. Writingb = b1/b2 with b1, b2 ∈ Z and gcd(b1, b2) = 1, we obtain

b22q = b2

1p

and so p | b2 and q | b1. Writing b1 = b′1q and b2 = b′2q for suitable b′1, b′2 ∈ Z, we obtain

(b′2)2p2q = (b′1)

2q2p,

hence(b′2)

2p = (b′1)2q.

From this we obtain p | b′1 and q | b′2; but then p | b1 as well as p | b2, contradicting the fact thatgcd(b1, b2) = 1. So

√q /∈ Q(

√p).

2.3. Arrange the induction carefully.

2.4. Notice that if v = ±u then b = v2 = u2 = a which is impossible; so v 6= ±u. Then

u− v =(u− v)(u + v)

u + v=

u2 − v2

u + v=

a− b

u + v∈ K(u + v).

Hence

u =12

((u + v) + (u− v)) ∈ K(u + v), v =12

((u + v)− (u− v)) ∈ K(u + v).

87

So K(u, v) 6 K(u + v) 6 K(u, v) and therefore K(u + v) = K(u, v).

2.5. Since 1, i span the Q-vector space Q(i), we have [Q(i) : Q] 6 2. But also if x, y ∈ R, thenx + yi = 0 ⇐⇒ x = y = 0, so 1, i is a basis for Q(i) over Q. Hence [Q(i) : Q] = 2.

2.6. First notice that [Q(√

3) : Q] = 2 (with Q-basis 1,√

3) and Q(√

3) 6 R. Also, i /∈ Q(√

3)and since i2 + 1 = 0, Q(

√3, i) = Q(

√3)(i) has [Q(

√3, i) : Q(

√3)] = 2. By Theorem 2.6(ii),

[Q(√

3, i) : Q] = [Q(√

3, i) : Q(√

3)] [Q(√

3) : Q] = 2× 2 = 4.

The following three subfields of Q(√

3, i) are distinct and are extensions of Q having degree 2:L1 = Q(

√3), L2 = Q(i), L3 = Q(

√3 i). Then [Lr ∩ Ls : Q] > 1 ⇐⇒ Lr ∩ Ls = Lr = Ls, so

Lr ∩ Ls = Q whenever r 6= s. The only real subfield amongst these is L1.

C2

∞R

∞ Q(√

3, i)2

ttttttttt2

2

KKKKKKKKKK

Q(√

3)

2 KKKKKKKKKKKQ(i)

2

Q(√

3 i)

2rrrrrrrrrrr

Q

2.7. (a) Since 5 is a prime,

[Q(ζ5) : Q] = [Q[X]/(Φ5(X)) : Q] = ϕ(5) = 5− 1 = 4.

(b) We have ζ5 = cos(2π/5)+sin(2π/5) i ∈ Q(ζ5). But also ζ−15 ∈ Q(ζ5) and ζ−1

5 = cos(2π/5)−sin(2π/5) i ∈ Q(ζ5). Hence we have

cos(2π/5) =12

(ζ5 + ζ−1

5

) ∈ Q(ζ5), sin(2π/5) i =12

(ζ5 − ζ−1

5

) ∈ Q(ζ5).

(c) This can be found by repeated use of the double angle formula

cos(A + B) = cos A cosB − sinA sinB.

The polynomial Tn(X) which expresses cosnt in terms of cos t is called the n-th Chebyshev poly-nomial, i.e., cos nt = Tn(cos t); the Maple command for computing this is orthopoly[T](n,X).In fact, Tn(X) ∈ Z[X].(d) For k = 0, 1, 2, 3, 4, cos(5(2kπ/5)) = cos(2kπ) = 1, so T5(cos 2kπ/5) − 1 = 0. So each ofthe numbers cos(2kπ/5) is a root of the polynomial T5(X)− 1 = (X − 1)(4X2 + 2X − 1)2. Fork = 1, 2, 3, 4, cos(2kπ/5) is a root of 4X2 + 2X − 1, hence

Q(cos(2π/5)) ∼= Q[X]/(4X2 + 2X − 1), [Q(cos(2kπ/5)) : Q] = 2.

(e) Q(ζ5)

2

Q(cos(2π/5))

2

Q

88 SOLUTIONS

2.8. This is similar to the previous question.


3.1. Clearly, t is algebraic over K if and only if ker εt 6= (0), i.e., (i) ⇐⇒ (ii). By Theorem 2.9,(ii) ⇐⇒ (iii). Hence these three conditions are indeed equivalent.

3.2. The diagrams at the bottom indicate useful subfields of the splitting fields occurring ineach of these examples.

p1(X) = X4−X2 +1: The polynomial X2−X +1 has the complex roots e±πi/3 =1±√3 i

2, so

the four roots of p1(X) are the complex square roots of these numbers, i.e., ±e±πi/6; explicitlythese are √

32

+12i, −

√3

2− 1

2i,

√3

2− 1

2i, −

√3

2+

12i.

The splitting field is E = Q(√

3, i) and [E : Q] = 4.p2(X) = X6 − 2: The roots are the six complex 6-th roots of 2, i.e., 6

√2e2kπi/6 with k =

0, 1, 2, 3, 4, 5; explicitly these are

6√

2,6√

2√

32

+6√

22

i, −6√

2√

32

+6√

22

i, − 6√

2, −6√

2√

32

−6√

22

i,6√

2√

32

−6√

22

i.

The splitting field is E = Q( 6√

2,√

3, i) and [E : Q] = 24.p3(X) = X4 + 2: The roots are the four 4-th roots of −2, 4

√2e(2k+1)πi/4 with k = 0, 1, 2, 3;

explicitly these are14√

2+

14√

2i, − 1

4√

2+

14√

2i, − 1

4√

2− 1

4√

2i,

14√

2− 1

4√

2i.

The splitting field is E = Q( 4√

2, i) and [E : Q] = 8.p4(X) = X4 + 5X3 + 10X2 + 10X + 5: Notice that

p4(Y − 1) = Y 4 + Y 3 + Y 2 + Y + 1 = Φ5(Y ),

so the splitting field of p4(X) over Q is the same as that of Φ5(Y ) over Q and this is thecyclotomic field Q(ζ5) where ζ5 = cos(2π/5) + sin(2π/5)i with [Q(ζ5) : Q] = 4; in fact, Q(ζ5) =Q(cos(2π/5), sin(2π/5)i).

Q(√

3, i)

Q(√

3)

2

Q

2

Q( 6√

2,√

3, i)

Q( 6√

2,√

3)

2

Q(√

3)

6

Q

2

Q( 4√

2, i)

Q( 4√

2)

2

Q

4

Q(cos(2π/5), sin(2π/5)i)

Q(cos(2π/5))

2

Q

2

3.3. List the three roots of X3−2 as u1 = 3√

2, u2 = 3√

2ζ3, u3 = 3√

2ζ23 . Then each automorphism

α ∈ AutQ(Q( 3√

2, ζ3)) permutes these roots, so can be identified with the unique permutationσα ∈ S3 for which

α(ui) = uσα(i) (i = 1, 2, 3).

We find that (using cycle notation for permutations)

σid = id, σα0 = (2 3), σα1 = (1 2 3), σα′1 = (1 2), σα2 = (1 3 2), σα′2 = (1 3).

89

These are the six elements of S3, therefore AutQ(Q( 3√

2, ζ3)) ∼= S3.

3.4. Irreducibility is a consequence of the polynomial version of the Eisenstein Criterion 1.43.Suppose that t ∈ k(T ) is a root of g(X); then using the Idiot’s Binomial Theorem we have

(X − t)p = Xp − tp = Xp − T,

so t is in fact a root of multiplicity p, hence is the only root in k(T ). This also gives thefactorization of g(X) into linear factors over k(T ).

3.5. Q(√

5,√

10)/Q: Here [Q(√

5,√

10) : Q] = 4 and the element√

5 +√

10 has degree 4 withminimal polynomial X4 − 30X2 + 25 which has roots ±√5±√10.Q(√

2, i)/Q: Here [Q(√

2, i) : Q] = 4 and the element√

2 + i has degree 4 with minimalpolynomial X4 − 2X2 + 9 which has roots ±√2± i.Q(√

3, i)/Q: Here [Q(√

3, i) : Q] = 4 and the element√

3 + i has degree 4 with minimalpolynomial X4 − 4X2 + 16 which has roots ±√3± i.Q( 4√

3, i)/Q: Here [Q( 4√

3, i) : Q] = 4 and the element 4√

3 + i has degree 8 with minimalpolynomial X8 + 4X6 + 40X2 + 4 which has roots ± 4

√3± i and ± 4

√3i± i.

3.6. The induction is straightforward. Here is the argument that K(u, v)/K is simple. Weassume that K is infinite since otherwise the result will be proved in Proposition 5.16.

Consider the subfields K(u + tv) 6 K(u, v) with t ∈ K. Then there are only finitely manyof these, so there must be s, t ∈ K such that s 6= t and K(u+ sv) = K(u+ tv). Then (s− t)v =(u+sv)−(u+tv) ∈ K(u+tv), hence v ∈ K(u+tv). This implies that u = (u+tv)−tv ∈ K(u+tv),hence K(u, v) 6 K(u + tv) 6 K(u, v) and therefore K(u, v) = K(u + tv).

3.7. If E/K is a quadratic extension then for any u ∈ E − K we have 1 < [K(u) : K] 6 2,so [K(u) : K] = 2 = [E : K] and therefore K(u) = E. Then minpolyK,u(X) must factor intolinear factors over E, so both its roots in K lie in E. This shows that E is normal over K.

The example F2(Z)/F2(Z2) is not separable since X2 − Z2 ∈ F2(Z2)[X] is irreducible butnot separable (see Qu. 3.4). If charK 6= 2 then all quadratic polynomials over K are separable.

3.8. Let E 6 C be a splitting subfield for f(X) over Q. Then if v ∈ C is a non-real root off(X) we have v /∈ Q(u), so f(X) does not split over Q(u) even though it has a root in this field.This means that there is a monomorphism ϕ ∈ MonoQ(Q(u),C) = MonoQ(Q(u),Q) for whichϕ(u) = v, hence ϕQ(u) 6= Q(u) and so Q(u)/Q is not normal.


4.1. By Proposition 3.81 we know that splitting fields are always normal, so it is only necessaryto show that the splitting field E of p(X) over K is separable over K. Since E is obtained byrepeatedly adjoining roots of p(X), the result follows from Proposition 3.73 together with thefact that if L/K 6 E/K is separable and v ∈ E is a root of p(X), then L(v)/K is separable.

4.2. (a) Suppose that f(X) = c3X3 + c2X

2 + c1X + c4 with c3 6= 0. Then

f(uX + v) =

c3u3X3 + (3c3vu2 + c2u

2)X2 + (3c3uv2 + c1u + 2c2uv)X + (c3v3 + c4 + c1v + c2v

2),

so if we take u to be any cube root of c3 and u = −c2/3c3 then f(uX + v) has the desired form.Notice that v ∈ K(u) and then f(uX + v) ∈ K(u), so provided that we can find a cube root of1/c3 in K, we have f(uX + v) ∈ K.(b) Viewing Gal(E/K) as a subgroup of S3, by Theorem 4.7 we know that 3 divides |Gal(E/K)|;but the only subgroups of S3 with this property are S3 and A3.(c) This is a tedious calculation! See Section 6.7 for the rest of this question.

90 SOLUTIONS

4.3. (a) This should be a familiar result.(b) The centre of D8 is

⟨α2

⟩which has order 2, and there are three normal subgroups of order

4, namely

〈α〉 = {ι, α, α2, α3}, ⟨α2, β

⟩= {ι, α2, β, βα2}, ⟨

α2, βα⟩

= {ι, α2, βα, βα3}.Notice that there are also four non-normal subgroups of order 2,

〈β〉 = {ι, β}, 〈βα〉 = {ι, βα}, ⟨βα2

⟩= {ι, βα2}, ⟨

βα3⟩

= {ι, βα3}.

4.4. This is an example of case (iii) of Kaplansky’s Theorem and we use the notation of theproof. The discriminant here is δ2 = −12, so we can take δ = 2

√3i. The roots of X2 + 3 are

±√3i, so we may assume

u = 4√

3 ζ8 =√

2 4√

32

(1 + i), v = 4√

3 ζ−18 =

√2 4√

32

(1− i),

where as usual ζ8 = e2πi/8 = (1 + i)/√

2. Hence we have uv =√

3 and uvδ = 6i. This gives thediagram of subfields of E

E = Q( 4√

3 ζ8,4√

3 ζ−18 ) = Q( 4

√3, ζ8)

Q(√

3, i)

jjjjjjjjjjjjjjjjjjj

TTTTTTTTTTTTTTTTTTT

Q(√

3i)

2UUUUUUUUUUUUUUUUUUUUUU Q(√

3)

2

Q(i)

2jjjjjjjjjjjjjjjjjjjjjj

Q

Then α is the restriction of complex conjugation to E, while β(√

3i) =√

3i and β(√

3) = −√3,hence also β(i) = −i. Using the choices of the proof, we have

β( 4√

3 ζ8) = − 4√

3 ζ8, β( 4√

3 ζ−18 ) = β(− 4

√3 ζ8i) = − 4

√3 ζ8i.

The effects of σ and γ on the four roots 4√

3 ζ8,4√

3 ζ−18 ,− 4

√3 ζ8,− 4

√3 ζ−1

8 of f(X) are given inpermutation notation by σ = (1 4 3 2) and α = (1 2)(3 4), and these generate a dihedralsubgroup of S4. Using the previous question (but beware that the notation there is inconsistentwith that of the present situation!) we have the normal subgroups

⟨σ2

⟩, 〈σ〉 , ⟨

σ2, α⟩,

⟨σ2, ασ

⟩,

and these have fixed fields

E〈σ2〉 = Q(√

3, i), E〈σ〉 = Q(i), E〈σ2,α〉 = Q(√

3), E〈σ2,ασ〉 = Q(√

3 i),

each of which is a normal extension of Q.

4.5. Q(X3 − 10)/Q: This is similar to Example 4.19, with splitting field Q( 3√

10, ζ3) andGal(Q( 3

√10, ζ3)/Q) ∼= S3.

Q(√

2)(X3 − 10)/Q(√

2): The splitting field is Q(√

2, 3√

10, ζ3). We have

Q(√

2,3√

10) 6 Q(√

2,3√

10, ζ3) 6 R

and [Q(√

2, 3√

10) : Q(√

2)] = 6. Since ζ3 is not real, we must have [Q(√

2, 3√

10, ζ3) : Q(√

2)] = 6.The Galois group is again isomorphic to S3.

91

Q(√

3 i)(X3 − 10)/Q(√

3 i): Here Q(√

3 i) = Q(ζ3), with [Q(ζ3) : Q] = 2. The splitting field isQ( 3√

10, ζ3) and [Q( 3√

10, ζ3) : Q(ζ3)] = 3, hence Gal(Q( 3√

10, ζ3)/Q(ζ3))Z/3 with generator σ forwhich σ( 3

√10) = 3

√10 ζ3.

Q(√

23 i)(X3 −X − 1)/Q(√

23 i): First note that X3 −X − 1 ∈ Z[X] must be irreducible sinceits reduction modulo 2 X3 + X + 1 ∈ F2[X] has no root in F2 and hence has no linear factor(see Qu. 1.10). To proceed further we can use the ideas of Qu. 4.2 above (see also Section 6.7).The discriminant of the polynomial X3 − X − 1 is ∆ = −23 and so δ =

√23 i. Then if

E = Q(√

23 i)(X3 − X − 1) is the splitting field of X3 − X − 1 over Q, Gal(E/Q) ∼= S3 andGal(E/Q(

√23 i)) ∼= A3.

K(X3 −X − 1)/K for K = Q, Q(√

5), Q(√

5 i), Q(i): Continuing the above discussion, noticethat [E ∩ R : Q] = 3, so

√5 /∈ E, hence

Q(√

5)(X3 −X − 1) = Q(X3 −X − 1)(√

5)

and [Q(√

5)(X3 − X − 1) : Q(√

5)] = [Q(X3 − X − 1) : Q] = 6, hence Gal(Q(√

5)(X3 − X −1)/Q(

√5)) ∼= S3. Similarly,

√5 i /∈ E and i /∈ E, hence

Gal(Q(√

5 i)(X3 −X − 1)/Q(√

5 i)) ∼= S3∼= Gal(Q(i)(X3 −X − 1)/Q(i)).


5.1. By Theorem 1.13, an integral domain D always has a monomorphism into a field j : D −→F (e.g., F can be taken to be its field of fractions), so any subgroup U 6 D× becomes isomorphicto a subgroup jU 6 F×, and if U is finite so is jU , therefore jU and U are cyclic.

5.2. The only root of X2 + 1 in F2 is the multiple root 1.

5.3. The field Fpd [X]/(f(X)) is an extension of Fpd which has degree n, hence it is a finitefield with pdn elements, hence Proposition 5.6 implies that it is isomorphic to Fpdn . Since theextension Fpdn/Fpd is normal, Fpdn is a splitting field for f(X) over Fpd .

5.4. (a) Here 41 is prime. Since 8 | (41− 1), there is a primitive 8-th root of unity in F41. 6 isa primitive root for F41 and 65 ≡ 27 mod 41 has order 8.(b) Here 5 is prime 4 | (5− 1), so there is a primitive 4-th root of unity in F×5 , but no primitive8-th root of unity. In fact, 2 and 3 have order 4, so these are primitive roots for F5. Notice thatin F5[X],

X8 − 1 = (X4 − 1)(X4 + 1) = (X4 − 1)(X2 − 2)(X2 − 3),

where the polynomials X2−2 and X2−3 are irreducible. Therefore F25 is the splitting field forX8 − 1 over F5 and we have F25

∼= F5(u) = F5(v), where u2 = 2 and v2 = 3, so ±u and ±v areprimitive 8-th roots of unity. To find an element of order 24 in F×25, we first find one of order 3.Consider the polynomial X2 + X + 1 ∈ F5[X]; in F5, this has roots which have order 3. Theseroots are given by (−1± w)/2, where w2 = (1− 4) = −3 = 2, hence they are

(−1± u)2

= −3± 3u.

Now the elements ±(2± 2u)u = ±(±4 + 2u) = ±4± 2u all have order 8× 3 = 24.(c) Here 11 is prime and 8 | (121−1) = 120, so F121 is the splitting field of X8−1 over F11. Thepolynomial X2 +1 is irreducible over F11 so F121 = F11(z) where z2 = −1. Since 120 = 8×3×5,it is sufficient to find elements of order 8, 3 and 5 and then their product will have order 120.

Suppose that a+ bz ∈ F121 with a, b ∈ F11. If this element has order 8, then (a+ bz)2 = ±z.So let us solve

(a2 − b2) + 2abz = z.

92 SOLUTIONS

Then 2ab = 1 and b2 = a2, hence b = ±a. Now we have 2a2 = ±1 and so a2 = ±1/2 = ±6.Now 6 is not a square in F11 but

72 ≡ −6 ≡ 42 mod 11,

so we have a = 4, b = ±4 and a = 7, b = ±7. Therefore the elements of order 8 in F×121 are4± 4z and 7± 7z.

By the same approach as in (b), we find the elements of order 3 in F121 to be (−1±5z)/2 =5± 8z.

2 is a primitive root for F11 so 4 = 22 has order 5.Combining these we obtain the following primitive roots for F121: 7± z, 10± 4z.

(d) In F2[X] we have X8 − 1 = (X − 1)8, so the only root in F2 is 1, hence the splitting field isF2.

5.5. Notice that Fp(w) is a splitting field of the separable polynomial Xpd−1 − 1 over Fp, soif w ∈ F×

p` then Fp(w) 6 Fp` . Since Fp(w) = Fpd we have d 6 `; we also have degFpw = d.

The number of conjugates of w is d, hence each primitive root of Fpd has d conjugates and thetotal number of these is the number of generators of the cyclic group F×

pd∼= Z/(pd − 1), i.e.,

ϕ(pd − 1). Hence d | ϕ(pd − 1). This can also be interpreted in terms of the evident action ofGal(Fpd/Fp) ∼= Z/d on the set of all primitive roots of Fpd ; each orbit has exactly d elements,so the number of orbits is ϕ(pd − 1)/d which is an integer.

5.6. (a) First note that g′a(X) = −1, so ga(X) is separable, hence E/K is separable. If u ∈ E

is a root of ga(X), then for t ∈ Fpd ,

ga(u + t) = (u + t)pd − (u + t)− a = (upd − u− a) + (tpd − t) = (upd − u− a) = 0,

hence u+ t is also a root of ga(X). This means that E = K(u) since all the other roots of ga(X)lie in K(u). As ga(X) is irreducible over K, [E : K] = pd = |Gal(E/K)| and so the followingpd automorphisms are the elements of Gal(E/K):

σt : E −→ E; σt(u) = u + t (t ∈ Fpd).

It is easy to check that for s, t ∈ Fpd , σs◦σt = σs+t. Hence there is an isomorphism Gal(E/K) ∼=Fpd with σt corresponding to t ∈ Fpd .(b) If ga(X) is irreducible over K then it cannot have a root in K since p > 1.

Conversely, suppose that ga(X) has no root in K. Then if u ∈ E is any root of ga(X) ina splitting field over K, the other roots are the p elements u + t ∈ E (t ∈ Fp). If u + t0 6= u

is a conjugate of u with 0 6= t0 ∈ Fp, there must be an element τt0 ∈ Gal(E/K) for whichτt0(u) = u + t0. Then 〈τt0〉 must be isomorphic to a non-trivial subgroup of Fp, but this mustbe Fp since this group is simple. Hence, u must have p conjugates and so ga(X) is irreducibleover K.(c) If K is a finite field and d > 1 then if ga(X) were irreducible over K, then by (a), E wouldbe finite and Gal(E/K) ∼= Fpd . But Fpd is not cyclic, yet we know from Proposition 5.23 thatGal(Fpd/Fp) ∼= Z/d is cyclic.


6.1. Now when n = 1, G ∼= Z/p, which is abelian. Suppose that the result holds whenever|G| = pk with k < n. Now if |G| = pn, recall that by Cauchy’s Lemma, the centre Z of G

is non-trivial. Hence G/Z has order |G/Z| = pk with k < n. By the inductive hypothesis,there is a normal subgroup M / G/Z with |M | = pk−1. By one of the Isomorphism Theorems,there is a normal subgroup N / G containing Z and satisfying N/Z = M ⊆ G/Z. Clearly|N | = |Z| |M | = pn−1. This establishes the inductive step and hence the desired result.

93

6.2. In this situation, for any non-zero t ∈ K, −t 6= t (since otherwise 2t = 0 and so t = 0).If ζ ∈ K is a primitive n-th root of unity, then (−ζ)n = (−1)nζn = −1, while (−ζ)2n =(−1)2nζ2n = 1. Hence −ζ ∈ K is a primitive 2n-th root of unity.

6.3. (a) We have ϕ(24) = ϕ(8)ϕ(3) = 4 × 2 = 8. The elements of Z/24 which are invertibleare the residue classes modulo 24 of the numbers 1, 5, 7, 11, 13, 17, 19, 23. For each of thesenumbers r, the residue class modulo 24, r, satisfies r2 = 1, hence these all have order except 1which has order 1. Since (Z/24)× is abelian, it is isomorphic to Z/2×Z/2×Z/2. The effect ofthese elements on Q(ζ24) is given by r · ζr

24. Notice that 23 acts like complex conjugation. Theeffect on Q(cos(π/12)) is given by

r · cos(π/12) = cos(πr/12),

so in particular,

−r · cos(π/12) = cos(−πr/12) = cos(πr/12) = r · cos(π/12).

(b) This is similar to (a). We have ϕ(20) = ϕ(4)ϕ(5) = 2× 4 = 8 and the elements of (Z/20)×

are the residue classes modulo 20 of the numbers 1, 3, 7, 9, 11, 13, 17, 19. This time there areelements of order 4, for instance 7 and 13. Then we have (Z/20)× ∼= Z2× Z/4.

6.4. For any n > 1, let ζn = e2πi/n = cos(2π/n) + sin(2π/n) i. Notice that if n is odd, thenQ(ζn) = Q(−ζn) where −ζn is a primitive 2n-th root of unity, so we might as well assume thatn is even from now on. We also have

ζn − ζ−1n = 2 sin(2π/n) i ∈ Q(ζn).

(a) If 4 - n then writing n = 2k with k odd, we have

[Q(ζn) : Q] = ϕ(2k) = ϕ(2)ϕ(k) = ϕ(k),

while[Q(ζ2n) : Q] = ϕ(4k) = ϕ(4)ϕ(k) = 2ϕ(k).

Hence, Q(ζn) cannot contain ζ2n and by another simple argument it cannot contain i = ζk2n. So

we see that sin(2π/n) /∈ Q(ζn) in this situation. Notice that since i = ζk2n,

sin(2π/n) =ζ22n − ζ−2

2n

2i∈ Q(ζ2n),

and by Proposition 6.3,

sin(2π/n) ∈ Q(ζ2n) ∩ R = Q(cos(π/n)).

Also, we have[Q(cos(π/n)) : Q] = 2[Q(cos(2π/n)) : Q],

henceQ(cos(π/n)) = Q(cos(2π/n))(sin(2π/n))

and[Q(cos(2π/n))(sin(2π/n)) : Q(cos(2π/n))] = 2,

withminpolyQ(cos(2π/n)),sin(2π/n)(X) = X2 + cos2(2π/n)− 1.

If 4 | n, we can write n = 4`. Then i = ζ`n, so i ∈ Q(ζn), whence

sin(π/2`) = sin(2π/n) =ζn − ζ−1

n

i∈ Q(ζn).

Clearlysin(π/2`) ∈ Q(ζn) ∩ R = Q(cos(2π/n)).

94 SOLUTIONS

Consider the automorphism σ ∈ Gal(Q(ζn)/Q)) for which σ(ζn) = ζ2`+1n = −ζn; it is easy to

see that σ has order 2. Then

σ(cos(2π/n)) = σ(cos(π/2`)) = − cos(π/2`),

σ(cos(π/`)) = cos(π/`),

σ(sin(2π/n)) = σ(sin(π/2`)) =−ζn + ζ−1

n

2(−ζn)`=

{sin(π/2`) if ` is odd,

− sin(π/2`) if ` is even.

From this we find that when ` is odd,

Q(cos(2π/n)) = Q(cos(π/2`)) = Q(cos(π/`))(sin(π/2`)) = Q(sin(π/2`)),

since cos(π/`) = 1− 2 sin2(π/2`) ∈ Q(sin(π/2`)). Thus we have [Q(sin(π/2`)) : Q] = 2ϕ(`) and

Gal(Q(sin(π/2`))/Q) = Gal(Q(cos(π/2`))/Q) = (Z/4`)×/{1,−1}.Similarly, if ` is even,

[Q(cos(π/`))(sin(π/2`)) : Q(cos(π/`))] = 2

and we must haveQ(cos(2π/n)) = Q(cos(π/2`)) = Q(sin(π/2`))

withGal(Q(sin(π/2`))/Q) = Gal(Q(cos(π/2`))/Q) = (Z/4`)×/{1,−1}

(b) We have

sin2(π/12) =1− cos(π/6)

2=

2−√34

,

and so

sin(π/12) =

√2−√3

2=√

6−√24

.

ThenQ(sin(π/12)) = Q(

√6−

√2) = Q(

√2,√

3).

andGal(Q(sin(π/12))/Q)) ∼= (Z/4`)×/{1,−1} ∼= Z/2× Z/2.

Here the effect of the coset of the residue class of r ∈ (Z/4`)× is given by

r · sin(π/12) =ζr24 − ζr

24

ir= sin(rπ/12) i1−r.

Explicitly we have

1 · sin(π/12) = −1 · sin(π/12) = sin(π/12) =√

6−√24

,

5 · sin(π/12) = −5 · sin(π/12) = sin(5π/12) =√

6 +√

24

,

7 · sin(π/12) = −7 · sin(π/12) = − sin(7π/12) =−√6−√2

4,

11 · sin(π/12) = −11 · sin(π/12) = − sin(11π/12) =−√6 +

√2

4.

In terms of the generators√

2 and√

3 these act by

1 ·√

2 =√

2, 1 ·√

3 =√

3, 5 ·√

2 = −√

2, 5 ·√

3 =√

3,

7 ·√

2 =√

2, 5 ·√

3 = −√

3, 11 ·√

2 = −√

2, 11 ·√

3 = −√

3.

95

6.5. Write n = 2kpr11 · · · prs

s , where each pj is an odd prime, p1 < p2 < · · · < ps, rj > 1 andk > 0. Then

ϕ(n) = ϕ(2k)ϕ(pr11 ) · · ·ϕ(prs

s ) = ϕ(2k)(p1 − 1)pr1−11 · · · (ps − 1)prs−1

s .

If s > 0 then ϕ(n) | 4 happens precisely when r1 = · · · = rs = 1 and one of the followingpossibilities occurs:

• p1 = 5, s = 1 and k = 0 (hence n = 5);• p1 = 3, s = 1 and k = 0, 1, 2 (hence n = 3, 6, 12);• s = 0 and k = 0, 1, 3 (hence n = 1, 2, 4, 8).

Q(i): Here degree [Q(i) : Q] = 2 and clearly the four 4-th roots of unity ±1,±i lie in this field.As ϕ(5) = 4, it has no 5-th roots of unity except 1. If it contained a 3-rd root of unity then itwould contain

√3 and so Q(

√3, i) 6 Q(i) which is impossible since [Q(

√3, i) : Q] = 4. From

this we see that the only roots of unity in Q(i) are ±1,±i.Q(√

2): Since this field is real, the only roots of unity it contains are ±1.Q(√

2 i) and Q(√

3): These field contains only the two square roots of unity ±1.

Q(√

3 i): This contains the six 6-th roots of unity ±1,±12±√

32

i.

Q(√

5) and Q(√

5 i): These fields contain only the two square roots of unity ±1.

6.6. Recalling the well-known formula

σ(i1 · · · ir)σ−1 = (σ(i1) · · · σ(ir)),

for 1 6 r 6 n− 2 we have

(1 2 · · · n)r(1 2)(1 2 · · · n)n−r = (1 2 · · · n)r(1 2)((1 2 · · · n)r)−1 = (r + 1 r + 2),

while

(1 2 · · · n)n−1(1 2)((1 2 · · · n)n−1)−1 = (1 2 · · · n)−1(1 2)((1 2 · · · n)−1)−1 = (n 1) = (1 n).

This means that every such 2-cycle (r + 1 r + 2) is in H. Also recall that every permutationρ ∈ Sn is a product of 2-cycles, so it suffices to show that every 2-cycle (a b) ∈ Sn is a productof 2-cycles of the form (r + 1 r + 2). Assuming that a < b, we also have

(a b) = (b− 1 b) · · · (a + 2 a + 3)(a + 1 a + 2)(a a + 1)(a + 1 a + 2)(a + 2 a + 3) · · · (b− 1 b),

and this is in H. Hence H = Sn.

6.7. This can be proved by induction on n. Write

e[m]r =

∑

i1<i2<···<ir6m

Xi1 · · ·Xir , s[m]r =

∑

16i6m

Xri .

Then we easily find that

e[m]r = e[m−1]

r + e[m−1]r−1 Xm, s[m]

r = s[m−1]r + Xr

m.

Notice also that e[m]r = 0 whenever r > m. The desired result is that for all n > 1 and k > 1,

s[n]k = e

[n]1 s

[n]k−1 − e

[n]2 s

[n]k−2 + · · ·+ (−1)k−1e

[n]k−1s

[n]1 + (−1)kke

[n]k .

96 SOLUTIONS

When n = 1 we have s[1]r = Xr

1 and e[1]1 = X1 from which the result follows. Now suppose

that the result is true for some n > 1. Then s[n+1]k = s

[n]k + Xk

n+1, while

e[n+1]1 s

[n+1]k−1 − e

[n+1]2 s

[n+1]k−2 + · · ·+ (−1)k−1e

[n+1]k−1 s

[n+1]1 + (−1)kke

[n+1]k =

(e[n]1 + Xn+1)(s

[n]k−1 + Xk−1

n+1)− (e[n]2 + e

[n]1 Xn+1)(s

[n]k−2 + Xk−2

n+1) + · · ·+ (−1)k−1(e[n]

k−1 + e[n]k−2Xn+1)(s

[n]1 + Xn+1) + (−1)kk(e[n]

k + e[n]k−1Xn+1)

= s[n]k + (e[n]

1 Xk−1n+1 − e

[n]2 Xk−2

n+1 + · · ·+ (−1)k−1e[n]k−1Xn+1)

+ (s[n]k−1 − e

[n]1 s

[n]k−2 + · · ·+ (−1)k−1e

[n]k−2s

[n]1 + (−1)kke

[n]k−1)Xn+1

+ (Xkn+1 − e

[n]1 Xk−1

n+1 + · · ·+ (−1)k−1e[n]k−2X

2n+1)

= s[n]k + Xk

n+1 = s[n+1]k ,

which demonstrates the inductive step.

6.8. (a) We have h1 = e1, h2 = e21 − e2 and h3 = e3 − 2e1e2 + e3

1.(b) can be done by induction on n in a similar way to the previous question.

6.9. (a) For each u ∈ E,

σ(u + σ(u) + σ2(u) + · · ·+ σn−1(u)) = σ(u) + σ2(u) + · · ·+ σn(u)

= σ(u) + σ2(u) + · · ·+ σn−1(u) + u

= u + σ(u) + σ2(u) + · · ·+ σn−1(u),

so the latter element is fixed by σ and all its powers, hence it is in K. It is straightforward toverify that the resulting function TrE/K : E −→ K is K-linear.(b) Let v ∈ E and suppose that TrE/K(v) = 0. By Artin’s Theorem 6.14, the linear combinationof characters id +σ + · · · + σn−1 cannot be linearly independent, so there is an element t ∈ E

for whichTrE/K t = t + σ(t) + · · ·+ σn−1(t) 6= 0.

Thenu = vσ(t) + (v + σ(v))σ2(t) + · · ·+ (v + σ(v)σ2(t) + · · ·+ σn−2(v))σn−1(t)

satisfies

u− σ(u) = v(σ(t) + σ2(t) + · · ·+ σn−1(t)

)− (σ(v) + · · ·+ σn−1(v)

)t

= v(t + σ(t) + σ2(t) + · · ·+ σn−1(t)

)− (v + σ(v) + · · ·+ σn−1(v)

)t

= (TrE/K t)v − (TrE/K v)t = (TrE/K t)v.

So we obtain

v =1

TrE/K tu− σ

(1

TrE/K tu

).

Notes for 4H Galois Theory 2001–2 Andrew...

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